Given an integral $I(x) = \int_a^b F(t, x) dt$, where $a$ and $b$ are constants, the rule is:

$\frac{dI}{dx} = \int_a^b \frac{\partial F}{\partial x}(t, x) dt$

This manipulation is valid if $F(t, x)$ and its partial derivative $\frac{\partial F}{\partial x}$ are jointly continuous in the rectangle defined by $a \le t \le b$ and the relevant range of $x$.

1. Dominance: There exists a function $k(t)$, which depends only on $t$, such that $|K(t, x)| \le k(t)$ for all $t \ge a$ and for all $x$ in the interval $[c, d]$.

2. Convergence: The improper integral $\int_a^\infty k(t) dt$ is convergent.

For an integral $I(x) = \int_{p(x)}^{q(x)} F(t, x) dt$, the Leibniz rule is:

$\frac{dI}{dx} = F(q(x), x)q'(x) - F(p(x), x)p'(x) + \int_{p(x)}^{q(x)} \frac{\partial F}{\partial x}(t, x) dt$

This rule is valid if $F$ and $\frac{\partial F}{\partial x}$ are jointly continuous, and $p(x)$ and $q(x)$ are differentiable.

We can rewrite the function using the identity $x^t = e^{t \ln x}$.
1. The function $\ln x$ is continuous for $x > 0$. We can treat it as a jointly continuous function of $(t, x)$.
2. The function $t$ is continuous for all $t \in \mathbb{R}$. We can also treat it as a jointly continuous function.
3. The product of two jointly continuous functions, $t \ln x$, is also jointly continuous.
4. The exponential function $e^y$ is continuous for all $y$.
5. Since $x^t$ is a composition of continuous functions, $e^y$ and $y = t \ln x$, it is also jointly continuous for all points where it is defined ($x>0, t \in \mathbb{R}$).

Dominated convergence ensures that the 'tails' of the improper integrals behave uniformly. When differentiating $\int_a^\infty F(t,x) dt$, we are concerned with the limit of $\int_a^T \frac{\partial F}{\partial x} dt$ as $T \to \infty$. Dominated convergence of $\frac{\partial F}{\partial x}$ guarantees that this limit exists and is independent of the parameter $x$, which allows the interchange to be valid. Without it, the convergence of the integral of the derivative might depend on $x$ in a way that invalidates the operation.

We differentiate under the integral sign with respect to $x$.
$I'(x) = \int_0^\pi \frac{\cos t}{1 + x \cos t} dt$.
We can rewrite the integrand as $\frac{1}{x} \left(1 - \frac{1}{1 + x \cos t}\right)$.
$I'(x) = \frac{1}{x} \int_0^\pi \left(1 - \frac{1}{1 + x \cos t}\right) dt = \frac{1}{x} \left[ t - \frac{\pi}{\sqrt{1-x^2}} \right]_0^\pi = \frac{\pi}{x} \left(1 - \frac{1}{\sqrt{1-x^2}}\right)$.
(The integral $\int \frac{dt}{1+a \cos t}$ is a standard result).

A proper integral $\int_a^b f(x) dx$ has a finite interval of integration $[a, b]$ and an integrand $f(x)$ that is bounded on that interval.

An improper integral is an integral where one or both of these conditions are not met. There are two main kinds:
1. Type 1: The interval of integration is infinite. Example: $\int_1^\infty \frac{1}{x^2} dx$.
2. Type 2: The integrand has a finite or infinite discontinuity at some point within the interval of integration. Example: $\int_0^1 \frac{1}{\sqrt{x}} dx$.

We need to find a dominating function $k(t)$. First, we find the maximum value of $|K(t,x)| = te^{-t/x}$ with respect to $x$ for a fixed $t$.
The partial derivative with respect to $x$ is $\frac{\partial}{\partial x}(te^{-t/x}) = t e^{-t/x} (\frac{t}{x^2}) > 0$.
Since the derivative is positive, the function is increasing with $x$. Therefore, its maximum value on $[c, d]$ occurs at $x=d$.
So, $|K(t,x)| \le te^{-t/d}$. We can choose $k(t) = te^{-t/d}$.
The integral $\int_0^\infty te^{-t/d} dt$ converges (it evaluates to $d^2$).
Since both conditions are met, the function has dominated convergence.

If $f(x, y)$ is an integrable (in particular, jointly continuous) function over a rectangle $R = [a, b] \times [c, d]$, then the double integral can be evaluated as an iterated integral in either order:

$\iint_R f(x,y) dA = \int_c^d \left( \int_a^b f(x,y) dx \right) dy = \int_a^b \left( \int_c^d f(x,y) dy \right) dx$

For a proper double integral of a continuous function over a rectangular region, Fubini's theorem states that the order of integration can be switched without changing the result.

If $0 \le f(x) \le g(x)$ for all $x \ge a$, then:
1. If $\int_a^\infty g(x) dx$ converges, then $\int_a^\infty f(x) dx$ converges.
2. If $\int_a^\infty f(x) dx$ diverges, then $\int_a^\infty g(x) dx$ diverges.

If $f(x) > 0$ and $g(x) > 0$ for $x \ge a$, and $\lim_{x \to \infty} \frac{f(x)}{g(x)} = L$, where $0 < L < \infty$, then $\int_a^\infty f(x) dx$ and $\int_a^\infty g(x) dx$ either both converge or both diverge.

Let $I(\alpha) = \int_a^\infty F(x, \alpha) dx$. We can differentiate under the integral sign, i.e., $I'(\alpha) = \int_a^\infty \frac{\partial F}{\partial \alpha} dx$, if $F(x, \alpha)$ is jointly continuous, the integral for $I(\alpha)$ converges, and the integral of $\frac{\partial F}{\partial \alpha}$ has dominated convergence.

The Dominated Convergence Theorem provides conditions for interchanging the limit and integral operations. If a sequence of functions $f_n(x)$ converges pointwise to $f(x)$ and is dominated by an integrable function $g(x)$ (i.e., $|f_n(x)| \le g(x)$ for all n), then $\lim_{n \to \infty} \int f_n(x) dx = \int \lim_{n \to \infty} f_n(x) dx = \int f(x) dx$.

If $g(x)$ is continuous at $x_0$ and $h(y)$ is continuous at $y_0$, then their product $F(x,y) = g(x)h(y)$ is jointly continuous at $(x_0, y_0)$. This is because the product of continuous functions is continuous.

Using the Leibniz rule with $F(t,x) = e^{-t^2}$, $q(x)=x^2$, $p(x)=x$:
$\frac{d}{dx} \int_x^{x^2} e^{-t^2} dt = e^{-(x^2)^2} \cdot (2x) - e^{-x^2} \cdot (1) + \int_x^{x^2} 0 dt$
$= 2xe^{-x^4} - e^{-x^2}$.

Yes. We test for absolute convergence by considering $\int_1^\infty |\frac{\sin x}{x^2}| dx$.
We know that $|\sin x| \le 1$, so $|\frac{\sin x}{x^2}| \le \frac{1}{x^2}$.
The integral $\int_1^\infty \frac{1}{x^2} dx$ is a p-integral with $p=2 > 1$, so it converges.
By the Direct Comparison Test, $\int_1^\infty |\frac{\sin x}{x^2}| dx$ converges. Therefore, the original integral is absolutely convergent.

Pointwise convergence: For a sequence of functions {$f_n(x)$}, for every $\epsilon > 0$ and for each $x$ in the domain, there exists an $N$ (which can depend on both $\epsilon$ and $x$) such that $|f_n(x) - f(x)| < \epsilon$ for all $n > N$.

Uniform convergence: For every $\epsilon > 0$, there exists an $N$ (which depends only on $\epsilon$) such that for all $x$ in the domain, $|f_n(x) - f(x)| < \epsilon$ for all $n > N$. The same $N$ works for all $x$.

Pointwise convergence of an integral $\int_a^\infty K(t,x) dt$ means that for any given $x$, the integral converges. However, the rate of convergence (how large $T$ must be for the tail $\int_T^\infty$ to be small) can depend on $x$.

Dominated convergence provides a single bounding function $k(t)$ whose convergent integral 'dominates' the convergence of all the integrals for every $x$ in a given set. This ensures a uniform rate of convergence for the tails of the integrals, independent of $x$.

False. A function can be continuous at a point but not have partial derivatives. For example, a cone-like surface $z = \sqrt{x^2+y^2}$ is continuous at the origin, but its partial derivatives are not defined there.

False. An integral can be conditionally convergent. This means the integral itself converges, but the integral of its absolute value diverges. A classic example is $\int_1^\infty \frac{\sin x}{x} dx$.

This is a Frullani integral. We can evaluate it by writing the integrand as $\int_a^b e^{-xy} dy$.
$\int_0^\infty \int_a^b e^{-xy} dy dx$. We can swap the order of integration (justification needed).
$\int_a^b \int_0^\infty e^{-xy} dx dy = \int_a^b [-\frac{1}{y}e^{-xy}]_0^\infty dy = \int_a^b \frac{1}{y} dy = [\ln y]_a^b = \ln(b/a)$.

Use integration by parts on $\int_0^\infty \frac{\sin^2 x}{x^2} dx$.
Let $u = \sin^2 x$ and $dv = \frac{1}{x^2} dx$. Then $du = 2 \sin x \cos x dx = \sin(2x) dx$ and $v = -\frac{1}{x}$.
The integral becomes $[-\frac{\sin^2 x}{x}]_0^\infty + \int_0^\infty \frac{\sin(2x)}{x} dx$.
The first term is 0. The second integral is a known result. Let $2x=t$, then $\int_0^\infty \frac{\sin t}{t/2} \frac{dt}{2} = \int_0^\infty \frac{\sin t}{t} dt = \frac{\pi}{2}$.

To find a dominating function $k(t)$, we maximize $|K(t,x)|$ with respect to $x$. The derivative w.r.t. $x$ is $\frac{\partial}{\partial x}(te^{-t/x}) = te^{-t/x}(\frac{t}{x^2}) > 0$. The function is increasing in $x$, so its maximum on $[c,d]$ is at $x=d$. Thus, $|K(t,x)| \le te^{-t/d}$. Let $k(t) = te^{-t/d}$. The integral $\int_0^\infty k(t) dt$ converges. Both conditions for dominated convergence are met.

This is an improper integral of the second kind because the integrand $\frac{1}{x}$ has a vertical asymptote (an infinite discontinuity) at $x=0$, which is the lower limit of integration.
To evaluate it, we compute $\lim_{a \to 0^+} \int_a^1 \frac{1}{x} dx = \lim_{a \to 0^+} [\ln x]_a^1 = \lim_{a \to 0^+} (\ln 1 - \ln a) = \lim_{a \to 0^+} (-\ln a) = \infty$.
Since the limit is infinite, the integral diverges.

This is a p-integral. It converges if $p > 1$ and diverges if $p \le 1$.
Proof:
If $p \ne 1$, $\int_1^b \frac{1}{x^p} dx = [\frac{x^{-p+1}}{-p+1}]_1^b = \frac{b^{1-p}-1}{1-p}$. As $b \to \infty$, this limit is finite (specifically $\frac{1}{p-1}$) only if $1-p < 0$, i.e., $p > 1$.
If $p=1$, $\int_1^b \frac{1}{x} dx = [\ln x]_1^b = \ln b$, which diverges as $b \to \infty$.

Fubini's Theorem allows the interchange of the order of integration for a double integral. A key condition for this theorem to hold is that the function $f(x,y)$ must be integrable over the rectangular domain. If the function is jointly continuous over a closed, bounded rectangle, its integrability is guaranteed, and therefore the order of integration can be validly swapped.

The function $f(x,y) = \frac{xy}{x^2+y^2}$ for $(x,y) \ne (0,0)$ and $f(0,0)=0$.
For any fixed $y_0$, $\lim_{x \to 0} f(x, y_0) = 0$, and for any fixed $x_0$, $\lim_{y \to 0} f(x_0, y) = 0$. So it is separately continuous.
However, approaching the origin along the line $y=mx$, the limit is $\lim_{x \to 0} \frac{mx^2}{x^2+m^2x^2} = \frac{m}{1+m^2}$. Since the limit depends on the path (the value of $m$), the function is not jointly continuous at (0,0).

The Jacobian of the transformation acts as a scaling factor for the differential area (or volume) element. When changing variables from $(x,y)$ to $(u,v)$, the area element $dx dy$ is replaced by $|J| du dv$, where $J$ is the Jacobian determinant of the transformation from $(u,v)$ to $(x,y)$.

$J = \det \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix}$

It accounts for the local distortion of area caused by the transformation.

You can swap the limit and the integral sign, i.e., $\lim_{x \to x_0} \int_a^b F(t,x) dt = \int_a^b \lim_{x \to x_0} F(t,x) dt$, if the integral is proper and $F(t,x)$ is jointly continuous in the rectangle $[a,b] \times [c,d]$ where $x_0$ is in the interval $[c,d]$.

For improper integrals, a stronger condition like dominated convergence is required.

A proper integral has a finite interval of integration and an integrand that is bounded over that interval.

An improper integral is one where either the interval of integration is infinite (e.g., $\int_1^\infty$) or the integrand is unbounded at one or more points within the interval (e.g., $\int_0^1 \frac{1}{x} dx$).

You must split the integral into two parts at the point of singularity. Then, you evaluate the limit for each part separately.

$\int_{-1}^1 \frac{1}{x^{1/3}} dx = \int_{-1}^0 \frac{1}{x^{1/3}} dx + \int_0^1 \frac{1}{x^{1/3}} dx$
$= \lim_{b \to 0^-} \int_{-1}^b x^{-1/3} dx + \lim_{a \to 0^+} \int_a^1 x^{-1/3} dx$
$= \lim_{b \to 0^-} [\frac{3}{2}x^{2/3}]_{-1}^b + \lim_{a \to 0^+} [\frac{3}{2}x^{2/3}]_a^1$
$= (0 - \frac{3}{2}) + (\frac{3}{2} - 0) = 0$.
Both integrals must converge for the original integral to converge.

The Cauchy Principal Value is a method of assigning a value to certain improper integrals that would otherwise be divergent. For an integral with a singularity at $x_0$ inside $(a,b)$, instead of taking two independent limits, it is defined as:

$P.V. \int_a^b f(x) dx = \lim_{\epsilon \to 0^+} \left( \int_a^{x_0-\epsilon} f(x) dx + \int_{x_0+\epsilon}^b f(x) dx \right)$

This symmetric approach to the singularity can result in a finite value even when the standard definition of the improper integral diverges.

We can split the integral: $\int_0^\infty e^{-x^2} dx = \int_0^1 e^{-x^2} dx + \int_1^\infty e^{-x^2} dx$.
The first part is a proper integral and is finite.
For the second part, note that for $x \ge 1$, we have $x^2 \ge x$, which implies $-x^2 \le -x$, and so $e^{-x^2} \le e^{-x}$.
The integral $\int_1^\infty e^{-x} dx$ converges (it equals $1/e$).
By the Direct Comparison Test, since $e^{-x^2}$ is bounded by a function whose integral converges, $\int_1^\infty e^{-x^2} dx$ also converges. Since both parts converge, the original integral converges.

An improper integral $\int_a^\infty f(x) dx$ is said to be absolutely convergent if the integral of the absolute value of the integrand, $\int_a^\infty |f(x)| dx$, converges.

An important theorem states that if an integral is absolutely convergent, then it is convergent.

An improper integral $\int_a^\infty f(x) dx$ is said to be conditionally convergent if the integral itself converges, but the integral of its absolute value, $\int_a^\infty |f(x)| dx$, diverges.

The integral $\int_1^\infty \frac{\sin x}{x} dx$.

The integral converges (can be shown using integration by parts).

However, the integral of its absolute value, $\int_1^\infty |\frac{\sin x}{x}| dx$, diverges. This can be shown by comparing it to a divergent series.

An improper integral $\int_a^\infty f(x, \alpha) dx$ converges uniformly for $\alpha$ in an interval $[c,d]$ if:
1. There exists a positive function $M(x)$ such that $|f(x, \alpha)| \le M(x)$ for all $\alpha$ in $[c,d]$.
2. The integral $\int_a^\infty M(x) dx$ converges.

This is a direct comparison test for uniform convergence.

Assuming we can differentiate under the integral sign:
$I'(a) = \int_0^\infty \frac{\partial}{\partial a} \frac{\sin(ax)}{x} dx = \int_0^\infty \frac{x \cos(ax)}{x} dx = \int_0^\infty \cos(ax) dx$.
This integral does not converge. This indicates that direct differentiation under the integral sign is not valid without further justification or a different approach. A different method is needed to evaluate the original integral.

The Gamma function is defined by the improper integral:

$\Gamma(x) = \int_0^\infty t^{x-1}e^{-t} dt$

This integral is convergent for $x > 0$.

The Beta function is defined by the integral:

$B(x, y) = \int_0^1 t^{x-1}(1-t)^{y-1} dt$

This integral is convergent for $x > 0$ and $y > 0$.

The Beta function can be expressed in terms of the Gamma function as follows:

$B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$

Using the property $\Gamma(n+1) = n!$ for positive integers $n$.

$\Gamma(5) = (5-1)! = 4! = 4 \times 3 \times 2 \times 1 = 24$.

Using the relationship $B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$:

$B(3, 4) = \frac{\Gamma(3)\Gamma(4)}{\Gamma(3+4)} = \frac{2! \cdot 3!}{6!} = \frac{2 \cdot 6}{720} = \frac{12}{720} = \frac{1}{60}$.

For large positive integers $n$, Stirling's formula provides an approximation for the factorial:

$n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n$

The integral, which defines the Beta function $B(x,y)$, is improper if the integrand is unbounded at the limits of integration.
1. At $t=0$, the term $t^{x-1}$ is unbounded if $x-1 < 0$, i.e., $x < 1$.
2. At $t=1$, the term $(1-t)^{y-1}$ is unbounded if $y-1 < 0$, i.e., $y < 1$.

Therefore, the integral is improper if $x < 1$ or $y < 1$. It still converges provided $x>0$ and $y>0$.

The value is $\Gamma(1/2) = \sqrt{\pi}$.

This can be shown by using the definition $\Gamma(1/2) = \int_0^\infty t^{-1/2}e^{-t} dt$ and making the substitution $t=u^2$, which transforms the integral into $2 \int_0^\infty e^{-u^2} du$. This is a form of the Gaussian integral, which evaluates to $\sqrt{\pi}$.

This technique is used to evaluate definite integrals. It involves introducing a parameter into the integrand, differentiating the integral with respect to that parameter (which often simplifies the integrand), evaluating the new, simpler integral, and then integrating the result with respect to the parameter to find the value of the original integral. It is a powerful method for integrals that are difficult to solve directly.

The Direct Comparison Test requires finding a comparison function that is strictly greater than (for convergence) or less than (for divergence) the original function over the entire tail of the integral. This can be difficult.

The Limit Comparison Test is often easier as it only requires finding a function that has the same 'dominant behavior' in the limit. You only need to show that the limit of the ratio of the two functions is a finite, non-zero constant, rather than satisfying a strict inequality.

Because the limit as $(x,y) \to (0,0)$ depends on the path of approach. If we approach along the line $y=mx$, the limit becomes:
$\lim_{x \to 0} \frac{x(mx)}{x^2 + (mx)^2} = \lim_{x \to 0} \frac{mx^2}{x^2(1+m^2)} = \frac{m}{1+m^2}$.
Since the value of the limit depends on $m$ (the slope of the path), a single limit does not exist, and thus the function is not jointly continuous at the origin.

Not necessarily. A function can be continuous, and even differentiable, without its derivative being continuous. For example, the function $f(x) = x^2 \sin(1/x)$ (with $f(0)=0$) is differentiable everywhere, but its derivative $f'(x)$ is not continuous at $x=0$. Similar examples exist for functions of two variables.

The Leibniz rule calculates the rate of change of an area (or volume) whose boundaries are moving. The total derivative $\frac{d}{dx} \int_{p(x)}^{q(x)} F(t,x) dt$ is the sum of three effects:
1. The change due to the top boundary moving: $F(q(x),x)q'(x)$.
2. The change due to the bottom boundary moving: $-F(p(x),x)p'(x)$.
3. The change due to the shape of the integrand function $F(t,x)$ changing with $x$: $\int_{p(x)}^{q(x)} \frac{\partial F}{\partial x} dt$.

To justify $\lim_{x \to x_0} \int_a^\infty F(t,x) dt = \int_a^\infty \lim_{x \to x_0} F(t,x) dt$, you need dominated convergence. If you can find a function $k(t)$ such that $|F(t,x)| \le k(t)$ for all $x$ in a neighborhood of $x_0$, and $\int_a^\infty k(t) dt$ converges, then the interchange is valid. This is often called the Dominated Convergence Theorem.

First, integrate with respect to $y$:
$\int_0^x (x+y) dy = [xy + \frac{y^2}{2}]_0^x = x(x) + \frac{x^2}{2} - 0 = x^2 + \frac{x^2}{2} = \frac{3}{2}x^2$.

Now, integrate the result with respect to $x$:
$\int_0^1 \frac{3}{2}x^2 dx = [\frac{3}{2} \frac{x^3}{3}]_0^1 = [\frac{x^3}{2}]_0^1 = \frac{1}{2} - 0 = \frac{1}{2}$.

The region of integration is defined by $y \le x \le 1$ and $0 \le y \le 1$. This is a triangle with vertices at (0,0), (1,0), and (1,1).
To change the order, we fix $x$ first. The limits for $x$ are from 0 to 1. For a fixed $x$, $y$ goes from $0$ to $x$.
The integral becomes:
$\int_0^1 \int_0^x e^{-x^2} dy dx$.

Fubini's theorem provides conditions under which it is possible to compute a double integral by using an iterated integral. For a function $f(x,y)$ that is integrable on a rectangle $R = [a,b] \times [c,d]$, the theorem states that:
$\iint_R f(x,y) dA = \int_c^d \int_a^b f(x,y) dx dy = \int_a^b \int_c^d f(x,y) dy dx$.
The theorem can be extended to more general regions as well.

A sufficient condition for Fubini's theorem to apply is that the function $f(x,y)$ is continuous on a closed and bounded (compact) region of integration. More generally, the theorem holds if the function is integrable, which is guaranteed if the set of discontinuities has measure zero. For improper integrals, the theorem (called the Fubini-Tonelli theorem) requires the function to be absolutely integrable.

The volume is given by the double integral $\int_0^1 \int_0^1 xy dx dy$.
$\int_0^1 [\frac{x^2y}{2}]_0^1 dy = \int_0^1 \frac{y}{2} dy = [\frac{y^2}{4}]_0^1 = \frac{1}{4}$.

The Riemann integral of $f(x,y)$ over a region $R$ is the limit of a Riemann sum. The region is partitioned into small sub-rectangles of area $\Delta A_i$. In each sub-rectangle, a sample point $(x_i^*, y_i^*)$ is chosen. The Riemann sum is $\sum_i f(x_i^*, y_i^*) \Delta A_i$. The integral is the limit of this sum as the size of the largest sub-rectangle goes to zero, provided the limit exists and is independent of the choice of sample points.

A Type I improper integral is one where the interval of integration is infinite. Examples include:
$\int_a^\infty f(x) dx$
$\int_{-\infty}^b f(x) dx$
$\int_{-\infty}^\infty f(x) dx$

A Type II improper integral is one where the integrand $f(x)$ has an infinite discontinuity at one or more points within a finite interval of integration $[a,b]$. For example, if $f(x)$ is discontinuous at $c \in [a,b]$, the integral is improper.

This is a p-integral with $p=3 > 1$, so it converges.
$\int_1^\infty \frac{1}{x^3} dx = \lim_{b \to \infty} \int_1^b x^{-3} dx = \lim_{b \to \infty} [\frac{x^{-2}}{-2}]_1^b$
$= \lim_{b \to \infty} (-\frac{1}{2b^2} - (-\frac{1}{2})) = 0 + \frac{1}{2} = \frac{1}{2}$.

Yes. This is a p-integral of the second kind, $\int_0^1 \frac{1}{x^p} dx$, with $p=1/2 < 1$. It converges.
Evaluation:
$\lim_{a \to 0^+} \int_a^1 x^{-1/2} dx = \lim_{a \to 0^+} [2x^{1/2}]_a^1 = \lim_{a \to 0^+} (2 - 2\sqrt{a}) = 2$.

No, it diverges. We can use the integral test or direct evaluation.
Evaluation:
Let $u = \ln x$, so $du = \frac{1}{x} dx$. The limits become $\ln 1 = 0$ and $\lim_{x \to \infty} \ln x = \infty$.
The integral becomes $\int_0^\infty u du = [\frac{u^2}{2}]_0^\infty$, which diverges.

For $x \ge 1$, we have $x^2+1 > x^2$, which implies $\frac{1}{x^2+1} < \frac{1}{x^2}$.
The integral $\int_1^\infty \frac{1}{x^2} dx$ is a p-integral with $p=2 > 1$, so it converges.
By the Direct Comparison Test, since the integrand is smaller than that of a convergent integral, $\int_1^\infty \frac{1}{x^2+1} dx$ also converges.

For large $x$, the integrand $\frac{x}{x^3+1}$ behaves like $\frac{x}{x^3} = \frac{1}{x^2}$. Let's compare it with $g(x) = \frac{1}{x^2}$.
$\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{x/(x^3+1)}{1/x^2} = \lim_{x \to \infty} \frac{x^3}{x^3+1} = 1$.
Since the limit is a finite non-zero constant, and since $\int_1^\infty \frac{1}{x^2} dx$ converges (p-integral with p=2), the original integral also converges.

The Laplace Transform of a function $f(t)$, denoted $\mathcal{L}\{f(t)\}$ or $F(s)$, is defined by the improper integral:

$F(s) = \int_0^\infty e^{-st} f(t) dt$

It is defined for all values of $s$ for which this integral converges.

$\mathcal{L}\{1\} = \int_0^\infty e^{-st} (1) dt = [-\frac{1}{s}e^{-st}]_0^\infty$
$= \lim_{b \to \infty} (-\frac{1}{s}e^{-sb}) - (-\frac{1}{s}e^0)$
$= 0 + \frac{1}{s} = \frac{1}{s}$, provided $s > 0$.

$\mathcal{L}\{e^{at}\} = \int_0^\infty e^{-st} e^{at} dt = \int_0^\infty e^{-(s-a)t} dt$
$= [-\frac{1}{s-a}e^{-(s-a)t}]_0^\infty$
$= 0 - (-\frac{1}{s-a}) = \frac{1}{s-a}$, provided $s > a$.

The Laplace transform of the derivative of a function $f(t)$ is given by:

$\mathcal{L}\{f'(t)\} = s\mathcal{L}\{f(t)\} - f(0) = sF(s) - f(0)$

1. Take the Laplace transform of the entire differential equation.
2. Use the properties of Laplace transforms (especially for derivatives) to convert the ODE into an algebraic equation in terms of $F(s)$.
3. Solve this algebraic equation for $F(s)$.
4. Find the inverse Laplace transform of $F(s)$ to get the solution $f(t)$.
This method incorporates initial conditions directly into the process.

The convolution of $f(t)$ and $g(t)$, denoted $(f * g)(t)$, is defined by the integral:

$(f * g)(t) = \int_0^t f(\tau)g(t-\tau) d\tau$

The Convolution Theorem states that the Laplace transform of the convolution of two functions is the product of their individual Laplace transforms:

$\mathcal{L}\{(f * g)(t)\} = \mathcal{L}\{f(t)\} \cdot \mathcal{L}\{g(t)\} = F(s)G(s)$

$I'(a) = \int_0^\infty \frac{-x \sin(ax)}{1+x^2} dx$.
$I''(a) = \int_0^\infty \frac{-x^2 \cos(ax)}{1+x^2} dx = \int_0^\infty \frac{(-(1+x^2)+1) \cos(ax)}{1+x^2} dx$
$= -\int_0^\infty \cos(ax) dx + \int_0^\infty \frac{\cos(ax)}{1+x^2} dx$.
The first integral diverges. This shows a problem with direct differentiation. A more careful approach is needed, often involving other methods to first establish convergence properties. However, if we formally proceed, the second term is $I(a)$. The invalidity of the first term suggests this direct approach is flawed. A correct proof uses different techniques.

If $g(t)$ is continuous at $t_0$ and $h(x)$ is continuous at $x_0$, then their sum $F(t,x)$ is jointly continuous at $(t_0, x_0)$. This is because the sum of continuous functions is continuous. Each function can be seen as a jointly continuous function of two variables where it is constant with respect to the other variable.

Yes. If $F(t,x)$ is jointly continuous on the rectangle $[a,b] \times [c,d]$, then the integral $I(x) = \int_a^b F(t,x) dt$ is a continuous function of $x$ on the interval $[c,d]$. This is a direct consequence of the rules for manipulating proper integrals (Rule 1 for limits).

Separate continuity means the function is continuous in each variable separately, holding the other constant. i.e., $f(x, y_0)$ is continuous in $x$ for fixed $y_0$, and $f(x_0, y)$ is continuous in $y$ for fixed $x_0$.

Joint continuity is a stronger condition. It means the function is continuous with respect to both variables simultaneously. The limit must exist and equal the function value as $(x,y)$ approaches $(x_0, y_0)$ along any path. A function can be separately continuous but not jointly continuous.

Using the Leibniz rule: $p(x)=0, q(x)=x, F(t,x)=\sin(xt)$.
$\frac{d}{dx} \int_0^x \sin(xt) dt = \sin(x \cdot x) \cdot (1) - \sin(x \cdot 0) \cdot (0) + \int_0^x \frac{\partial}{\partial x}(\sin(xt)) dt$
$= \sin(x^2) + \int_0^x t \cos(xt) dt$
$= \sin(x^2) + [\frac{t \sin(xt)}{x} + \frac{\cos(xt)}{x^2}]_0^x$
$= \sin(x^2) + (\frac{x \sin(x^2)}{x} + \frac{\cos(x^2)}{x^2}) - (0 + \frac{1}{x^2}) = 2\sin(x^2) + \frac{\cos(x^2)-1}{x^2}$.

No. Let's test the path $y=mx^2$.
$\lim_{x \to 0} \frac{x^2(mx^2)}{x^4+(mx^2)^2} = \lim_{x \to 0} \frac{mx^4}{x^4(1+m^2)} = \frac{m}{1+m^2}$.
Since the limit depends on the path (the value of $m$), the function is not jointly continuous at the origin.

We can write the integrand as $\int_b^a \frac{1}{1+(xy)^2} dy$.
The integral becomes $\int_0^\infty \int_b^a \frac{1}{1+(xy)^2} dy dx$.
Swapping the order of integration: $\int_b^a \int_0^\infty \frac{1}{1+(xy)^2} dx dy$.
The inner integral is $[\frac{1}{y} \arctan(xy)]_0^\infty = \frac{1}{y} \frac{\pi}{2}$.
The outer integral is $\int_b^a \frac{\pi}{2y} dy = [\frac{\pi}{2} \ln y]_b^a = \frac{\pi}{2}(\ln a - \ln b) = \frac{\pi}{2} \ln(a/b)$.

This is the real part of the Laplace transform of $e^{ibx}$.
$\int_0^\infty e^{-ax} e^{ibx} dx = \int_0^\infty e^{-(a-ib)x} dx = [-\frac{1}{a-ib}e^{-(a-ib)x}]_0^\infty = \frac{1}{a-ib} = \frac{a+ib}{a^2+b^2}$.
The real part is $\frac{a}{a^2+b^2}$.

This is the imaginary part of the Laplace transform of $e^{ibx}$.
$\int_0^\infty e^{-ax} e^{ibx} dx = \frac{a+ib}{a^2+b^2}$.
The imaginary part is $\frac{b}{a^2+b^2}$.

This is a famous result, often called the Basel problem. One method involves evaluating the double integral $\int_0^1 \int_0^1 \frac{1}{1-xy} dx dy$ in two different ways. One way leads to $\frac{\pi^2}{6}$, and the other way, by expanding the integrand as a geometric series $\sum (xy)^n$ and integrating term-by-term, leads to $\sum_{n=1}^\infty \frac{1}{n^2}$. Equating the two results proves the identity.

If $f(x)$ is a positive, continuous, and decreasing function for $x \ge N$, and if $a_n = f(n)$, then the series $\sum_{n=N}^\infty a_n$ and the improper integral $\int_N^\infty f(x) dx$ either both converge or both diverge.

This is a p-integral of the second kind with a singularity at the lower limit $a$.
The integral converges if $p < 1$ and diverges if $p \ge 1$.

This is a p-integral of the second kind with a singularity at the upper limit $b$.
The integral converges if $p < 1$ and diverges if $p \ge 1$.

We must split the integral at the singularity $x=0$:
$\int_{-1}^1 \frac{1}{x} dx = \int_{-1}^0 \frac{1}{x} dx + \int_0^1 \frac{1}{x} dx$.
The second integral is $\lim_{a \to 0^+} [\ln x]_a^1 = \lim_{a \to 0^+} (0 - \ln a) = \infty$.
Since one of the parts diverges, the entire integral diverges.

The Cauchy Principal Value is defined as:
$P.V. \int_{-1}^1 \frac{1}{x} dx = \lim_{\epsilon \to 0^+} \left( \int_{-1}^{-\epsilon} \frac{1}{x} dx + \int_{\epsilon}^1 \frac{1}{x} dx \right)$
$= \lim_{\epsilon \to 0^+} \left( [\ln|x|]_{-1}^{-\epsilon} + [\ln|x|]_{\epsilon}^1 \right)$
$= \lim_{\epsilon \to 0^+} (\ln\epsilon - \ln 1 + \ln 1 - \ln\epsilon) = \lim_{\epsilon \to 0^+} 0 = 0$.

The integral converges uniformly for $\alpha$ in a set $S$ if for every $\epsilon > 0$, there exists a number $N$ (depending only on $\epsilon$, not on $\alpha$) such that for all $\alpha \in S$:

$\left| \int_b^\infty f(x, \alpha) dx \right| < \epsilon$ for all $b > N$.

This means the tail of the integral can be made uniformly small for all $\alpha$ in the set.

The integral $I(\alpha) = \int_0^\infty \alpha e^{-\alpha x} dx$.
For any fixed $\alpha > 0$, the integral converges to 1. So it is pointwise convergent.
However, the convergence is not uniform on $(0, 1]$. To make the tail $\int_T^\infty \alpha e^{-\alpha x} dx = e^{-\alpha T}$ small, the choice of $T$ must depend on $\alpha$. As $\alpha \to 0^+$, we need an increasingly large $T$ to keep the tail small. No single $T$ works for all $\alpha$ in $(0,1]$.

If $f(x, \alpha)$ is a continuous function of both variables, and the improper integral $I(\alpha) = \int_a^\infty f(x, \alpha) dx$ converges uniformly for $\alpha$ in an interval, then $I(\alpha)$ is a continuous function of $\alpha$ in that interval. Uniform convergence allows us to say $\lim_{\alpha \to \alpha_0} I(\alpha) = I(\alpha_0)$.

If the integral $I(\alpha) = \int_a^\infty f(x, \alpha) dx$ converges, and the integral of the partial derivative, $\int_a^\infty \frac{\partial f}{\partial \alpha} dx$, converges uniformly, then we can differentiate under the integral sign: $I'(\alpha) = \int_a^\infty \frac{\partial f}{\partial \alpha} dx$. Uniform convergence of the derivative's integral is the key condition.

If the integral $I(\alpha) = \int_a^\infty f(x, \alpha) dx$ converges uniformly for $\alpha$ in $[c,d]$, then we can interchange the order of integration:
$\int_c^d I(\alpha) d\alpha = \int_c^d \left( \int_a^\infty f(x, \alpha) dx \right) d\alpha = \int_a^\infty \left( \int_c^d f(x, \alpha) d\alpha \right) dx$.

Using the property $\Gamma(x+1) = x\Gamma(x)$:
$\Gamma(3/2) = \Gamma(1/2 + 1) = \frac{1}{2}\Gamma(1/2)$.
Since $\Gamma(1/2) = \sqrt{\pi}$, we have $\Gamma(3/2) = \frac{\sqrt{\pi}}{2}$.

Using the Gamma function representation:
$B(x+1, y) = \frac{\Gamma(x+1)\Gamma(y)}{\Gamma(x+y+1)} = \frac{x\Gamma(x)\Gamma(y)}{(x+y)\Gamma(x+y)} = \frac{x}{x+y} B(x,y)$.

Let $u = \sin^2 x$. Then $du = 2 \sin x \cos x dx$. The integral can be transformed into a Beta function form.
The result is: $\frac{1}{2} B(\frac{m+1}{2}, \frac{n+1}{2})$.

The duplication formula, also known as Legendre's duplication formula, is:

$\Gamma(z) \Gamma(z + 1/2) = 2^{1-2z} \sqrt{\pi} \Gamma(2z)$

A Dirichlet integral is a type of multiple integral of the form:

$\iiint_V x^{a-1}y^{b-1}z^{c-1} dx dy dz$

over a region $V$ in the first octant bounded by the surface $(\frac{x}{p})^\alpha + (\frac{y}{q})^\beta + (\frac{z}{r})^\gamma = 1$ and the coordinate planes. These integrals have a known solution in terms of Gamma functions.

The transformation is $x = r \cos \theta$ and $y = r \sin \theta$.
The Jacobian determinant is:
$J = \det \begin{pmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{pmatrix} = \det \begin{pmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{pmatrix}$ $= (r \cos^2 \theta) - (-r \sin^2 \theta) = r(\cos^2 \theta + \sin^2 \theta) = r$.