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0 votes

Integral of exponential and modified bessel function of second kind

Using only Mathematica 13.1: $$\int_0^{\infty } \exp (-\alpha x) K_v\left(\beta \sqrt{x}\right) \, dx=-\frac{e^{\frac{\beta ^2}{8 \alpha }} \sqrt{\pi } \beta \left(K_{\frac{1}{2} (-1+v)}\left(\frac{...
1 vote

How to evaluate this improper integral $\int_{-\infty}^\infty x^2 e^{-x^2}\cos x \, dx$?

By successive integration by parts, we will show that $\int_{-\infty}^\infty x^2 e^{-x^2} \cos(x)\,dx=1/4\int_{-\infty}^\infty e^{-x^2} \cos(x)\,dx$. The latter integral can be evaluated a number of ...
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0 votes

Bochner integral: Is $f=g$ $\mu$-a.e. if their integrals are equal on every measurable set?

I fill some detail in @geetha290krm's excellent answer to better understand his/her ideas. We have $$ \varphi \left (\int_A f \mathrm d \mu\right ) = \varphi \left (\int_A g \mathrm d \mu \right) \...
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3 votes

How to evaluate this improper integral $\int_{-\infty}^\infty x^2 e^{-x^2}\cos x \, dx$?

How to solve! The actual result is different: $$ \int_{-\infty}^{\infty} x^{2} \cdot e^{-x^{2}} \cdot \cos(x) ~ \mathrm{d}x = \frac{\sqrt{\pi}}{4 \cdot \sqrt[4]{e}} \approx 0.345097 $$ The easiest way ...
3 votes

How to evaluate this improper integral $\int_{-\infty}^\infty x^2 e^{-x^2}\cos x \, dx$?

Consider the problem of the antiderivative $$I=\int x^2 e^{-x^2} \cos (x) \, dx=\Re \int x^2 e^{-x^2} e^{i x} \, dx=\Re \int x^2 e^{-x^2+ix} \, dx$$ Complete the square $$J=\int x^2 e^{-x^2+ix} \, ...
2 votes

If $\lim_{t\to +\infty} \int_{0}^{\pi} f(x)e^{xt} \, dx=0$ then $f=0$?

Remark Who knows the theorems about the rate that a Laplace transform goes to $0$? Is that relevant? In this problem, write $g(x) = f(\pi-x)$ for $x \in [0,\pi]$ and $g(x) = 0$ elsewhere. Then change ...
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4 votes

How to evaluate this improper integral $\int_{-\infty}^\infty x^2 e^{-x^2}\cos x \, dx$?

You could approach this with a recursive diffeq method. This is a bit inefficient but with enough effort this works. Let $$I(a)=\int^{\infty}_{-\infty}x^2 e^{-x^2}\cos(ax)\text{ d}x$$ Then \begin{...
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2 votes

Interchange of sum and integral (particular case).

The following two observations regarding power series will come handy: Suppose a power series $ f(x) = \sum_{n=0}^{\infty} a_n x^n$ has radius of convergence $R > 0$. By the Weierstrass M-test, ...
0 votes

Interchange of sum and integral (particular case).

Let $S_n(x)=x\sum_{k=0}^n(-x^3)^k$ and $S(x)=x\sum_{k=0}^\infty(-x^3)^k$. On $[0,1]$, $S_{2n-1}\le S\le S_{2n}$ and $\int_0^1(S_{2n}-S_{2n-1})=\frac1{6n+2}\to0$ hence $\int_0^1S=\lim\int_0^1S_n$.
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2 votes
Accepted

Bochner integral: Is $f=g$ $\mu$-a.e. if their integrals are equal on every measurable set?

If $f$ and $g$ are $\mu-$ integrable (in the Bochner sense ) then they are almost separably valued an this reduces the proof to the case when $E$ is a separable Banach Space. Now $\int_A x^{*}\circ f ...
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3 votes

Interchange of sum and integral (particular case).

For $x \in [0,1]$, we have $$0 \leqslant S_n(x) = \sum_{k=0}^n (-1)^k x^{3k+1} = x - (x^4 - x^7) - (x^9 - x^{13}) - \ldots \leqslant x$$ Since $x \mapsto x$ is integrable on $[0,1]$, the Dominated ...
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4 votes

Interchange of sum and integral (particular case).

The theorem is a sufficient condition but is not a necessary condition. To prove the exchange you can do this trick $$\int_0^1{x \sum_{n=0}^{\infty}{(-x^3)^n} dx} = \int_0^1{x\sum_{n=0}^{\infty}{(x^{...
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0 votes

Integration of density function

It is possible to do this integration directly, but it can be easier to choose a different change of variables. This can be a bit of an art, so I'll outline my thinking as I go. So whatever ...
4 votes

If $\lim_{t\to +\infty} \int_{0}^{\pi} f(x)e^{xt} \, dx=0$ then $f=0$?

It seems that $f$ is necessarily zero. Assume the converse and let $\text{esssup}f = [a,b]\subset[-\pi, \pi]$. Then the Fourier transform $F(z) = \int_a^b f(x) e^{izx}\,dx$ of $f$ is a entire function ...
0 votes

If $\lim_{t\to +\infty} \int_{0}^{\pi} f(x)e^{xt} \, dx=0$ then $f=0$?

$\newcommand{\d}{\,\mathrm{d}}$A partial response. Suppose $f\neq0\in L^2(0,\pi)$, that is, $\|f\|>0$. By Holder's inequality, for any $0<\eta<\pi$ and any $t>0$ we know: $$\left|\int_0^\...
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1 vote
Accepted

How to do the integral $\int_0^\pi\sin^{n}\theta\cos(a \cos\theta)\mathrm d\theta$?

Define $$F_n(a)=a^n\int_0^\pi\sin^{2n}x\cos(a\cos x)\,dx$$ Then differentiating on $a$ yields $$F_n'(a)=na^{n-1}\int_0^\pi\sin^{2n}x\cos(a\cos x)\,dx - a^n\int_0^\pi\sin^{2n}x\cos x\sin(a\cos x)\,dx$$ ...
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0 votes

How to evaluate $\int_{-\infty}^\infty x^n \frac{e^x}{(e^x+1)^2} dx$?

Suppose $n$ is a nonnegative integer. Clearly when $n$ is odd, $I_n=0$. For even $n$, one has \begin{eqnarray} I_n&=&\int_{-\infty}^\infty x^n \frac{e^x}{(e^x+1)^2} dx\\ &=&2\int_{0}^\...
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1 vote
Accepted

An IV problem using d'Alembert's solution: how does this integration work?

Generally, the name of a variable of integration doesn't actually matter, it's just an identifier. It only really matters when there's more than one variable involved that you could be integrating ...
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1 vote

Evaluate $\int \sqrt{x^2 + a^2}\mathrm{d}x$

W.l.o.g. we can assume that $a>0$ (because, otherwise one can replace $a$ with its absolute value) Setting $$ x=a\tan(u), $$ we have $$ dx=a\sec^2(u)du, $$ so we get \begin{eqnarray} \int\sqrt{x^2+...
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2 votes

Evaluate $\int \sqrt{x^2 + a^2}\mathrm{d}x$

Those types of integrals usually take advantage of some properties of the hyperbolic sine and cosine functions - $$\sinh t=\frac{e^t-e^{-t}}{2}\\\cosh t=\frac{e^t+e^{-t}}{2}\\(\sinh t)'=\cosh t\\(\...
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3 votes
Accepted

Evaluate $\int \sqrt{x^2 + a^2}\mathrm{d}x$

You have a little mistake in the integration by parts, $\displaystyle(\sqrt{x^2 + a^2})' = \frac{x}{x^2 + a^2}$ therefore $$ \int \sqrt{x^2 + a^2}\,dx = x\sqrt{x^2 + a^2} - \int \sqrt{x^2 + a^2}\,dx + ...
2 votes

How to evaluate $\int_{-\infty}^\infty x^n \frac{e^x}{(e^x+1)^2} dx$?

Too long for a comment $$I(n)=\int_{-\infty}^\infty x^n \frac{e^x}{(e^x+1)^2} dx=\int_{-\infty}^\infty \frac{x^n}{(e^{x/2}+e^{-x/2})^2} dx=2^{n-1}\int_{-\infty}^\infty \frac{t^n}{\cosh^2 t}dt$$ It is ...
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1 vote

How to evaluate $\int_{-\infty}^\infty x^n \frac{e^x}{(e^x+1)^2} dx$?

Here is an attempt using the Feynman trick. Consider the integral $$\int_{-\infty}^{\infty} \frac{e^{ax}}{(1+e^x)^2} \, \mathrm{d}x$$ for $a \in (0,2)$. Then $$\int_{-\infty}^{\infty} x^n\frac{e^{x}}{(...
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0 votes

Multiple integral problem (limits interchanged)

It comes from the traditional variable switch trick from multivariate calculus The region $$0 \leq x \leq L, 0 \leq \xi \leq x$$ is the same as the region $$0 \leq \xi \leq L, \xi \leq x \leq L$$
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2 votes

Multiple integral problem (limits interchanged)

Let's make a quick sketch of the domain of the integration. The horizontal axis is $x$, the vertical axis is $\xi$. In the first integral, $x$ goes from $0$ to $L$ (8 in the picture). The integration ...
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0 votes

Calculating $\int_0^1\frac{\ln^2x\ln(1-x)}{1-x}dx$ without using Beta function and Euler sum.

\begin{align}J&=\int_0^1 \frac{\ln^2 x\ln(1-x)}{1-x}dx\\ &\overset{\text{IBP}}=\underbrace{\left[\left(\int_0^x \frac{\ln^2 t}{1-t}dt-\int_0^1 \frac{\ln^2 t}{1-t}dt\right)\ln(1-x)\right]_0^1}_{...
  • 11.2k
1 vote

prove that $\lim\limits_{n\to\infty} x_n = \lim\limits_{n\to\infty} y_n$

Putting together (essentially) the proof from comments that the limits exist and are equal, by @AbhijeetVats and @DanielWainfleet: If $x=y$, then $x_n=y_n=x=y$ for all $n$, so obviously the limits ...
0 votes

Expected distance of a point from the center of a circle

$$ \frac 1{\pi R^2}\int_{x^2+y^2<R^2}\sqrt{x^2+y^2}dx dy = (polar\ coordinates) = \frac 1{\pi R^2}\int_{r<R}2\pi r^2 dr=\frac 23 R $$
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0 votes

Closed form of $\int_0^\infty \left(\frac{\arctan x}{x}\right)^ndx$

A complete asymptotic expansion for large $n$ follows from Laplace's method: \begin{align*} \int_0^{ + \infty } {\left( {\frac{{\arctan t}}{t}} \right)^n \mathrm{d}t} & = \int_0^{ + \infty } {\exp ...
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0 votes

Evaluate $\int_0^{\pi/2}\log\cos(x)\,\mathrm{d}x$

I first treat the integral as a derivative of a beta function $$ I(a)=\int_0^{\frac{\pi}{2}} \cos ^a x d x=\frac{1}{2} B\left(\frac{a+1}{2}, \frac{1}{2}\right) $$ $$ \begin{aligned} I&\left.=\frac{...
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1 vote

Prove that $\int_0^\infty\left(\arctan \frac1x\right)^2 \mathrm d x = \pi\ln 2$

Letting $y=\arctan \frac{1}{x} $ yields $$ \begin{aligned} I=& \int_0^{\frac{\pi}{2}} y^2 \csc ^2 y d y \\ &=-\int_0^{\frac{\pi}{2}} y^2 d(\cot y) \\ &=-\left[y^2 \cot y\right]_0^{\frac{\...
  • 7,419
0 votes

How to tackle the integral $\int_{0}^{\infty} \frac{\ln x}{x^{n}-1} d x$?

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} ...
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1 vote
Accepted

Integral representation of Bessel function of the first kind from generating function

There was a sign error in the integral representaiton of the first-kind Bessel function of order $n$ as stated in the OP. The correct representation of the Bessel function is $$J_{n}(x)=\frac1{2\pi}\...
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0 votes

Show that for non-negative measurable functions $f,g$ with $fg \geq 1$ the inequality $(\int f^p)(\int g^p) \geq 1$ holds.

Why not use C-B-S inequality $$\int f^p \cdot \int g^p \ge \left(\int \sqrt{f^p g^p}\right)^2 $$
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1 vote

How to show $ \int_{-\infty}^{\infty} \frac{e^{-(x+1)^2}}{1+e^{-x}}\mathrm{d}x = \frac{\left(2\sqrt[4]{e} -1 \right)\sqrt{\pi}}{2e}$?

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} ...
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1 vote

prove that $\lim\limits_{n\to\infty} x_n = \lim\limits_{n\to\infty} y_n$

Just showing that $x_1=x, y_1=y, x_{n+1}=\frac{x_n+y_n}{2}, y_{n+1}=\sqrt{x_ny_n}.$ Show that $\displaystyle \lim_{n\ \to \infty} x_n = \lim_{n \to \infty} y_n. $ let $x_n-y_n=d_n.$ Then, $d_n\geq 0$,...
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1 vote

Why is the volume of the solid of revolution between two graphs $\int_{0}^{b}\pi[f^2(x)-g^2(x)]dx$ instead of $\int_{0}^{b}\pi[f(x)-g(x)]^2dx$?

Consider a region bounded by the lines $x=0$ and $x=b$ and the curves $y=f(x)$ and $y=g(x)$, with $c\le g(x)\le f(x)$. Suppose we revolve this region about a line $y=c$ and use the washer method to ...
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2 votes

a question on improper integrals

If $f$ is positive then the exercise can be proved in the following way. Denote your function $x\mapsto \frac{1}{x}\int_0^x tf(t)\,dt$ by $g$. First of all, notice that $\displaystyle\lim_{x\to 0^+}g(...
6 votes
Accepted

Compute the definite integral $\int_{-1}^12\sqrt{4x^2+9x^4}dx$

The problem is that as you mentioned, $\sqrt{x^2} = |x| \ne x$, but there is an easy remedy: you can split the integral domain to 2 integrals over $(-1,0)$ and $(0,2)$, and then your technique will ...
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2 votes

Calculating $\int_{e^2}^{e^3} \frac{1}{x \ln(x)\ln(\ln(x))} dx$

Note $\frac{d}{dx}\left(\ln(\ln(x))\right)=\frac{1}{x\ln(x)}$ Let $u=\ln(\ln(x))$ and $du=\frac{1}{x\ln(x)}dx$ Bounds go to $\ln 2$ and $\ln 3$ $$\implies\int_{\ln 2}^{\ln{3}}\frac{1}{u}\;du$$ Can you ...
0 votes

integral with hyperspherical coordinates?

Though it does not have the direct relation to the formulated task, it would be interesting to get a closed form of $\displaystyle I(\xi,n)=\int_{\mathbb{R}^n} \frac{{\lvert e^{i\langle\vec\xi,\vec y\...
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1 vote
Accepted

Calculating $\int_{e^2}^{e^3} \frac{1}{x \ln(x)\ln(\ln(x))} dx$

This is due to rote substitution. Substitute $\color{red}{t=\ln(x)}$ and $\color{blue}{dt=\frac{dx}x}$. The limits change to $x=e^2\implies t=\ln(e^2)=2$ and $x=e^3\implies t=\ln(e^3)=3$. $$\int_{x=e^...
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-1 votes

Simplify $f\Big(\text{argmax}_{y\in \mathcal{Y}} \int_{\mathcal{X}} f(y,x) g(x) dx, x\Big)$

Perhaps the analog is: $$ \int_{\mathcal X}f\Big(\operatorname{argmax}_{y\in \mathcal{Y}} \int_{\mathcal{X}} f(y,t) g(t) dt,\; x\Big)\;dx = \max_{y\in \mathcal{Y}} \int_{\mathcal{X}} f(y,t) g(t) dt $$
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1 vote

Calculate $ϕ(\lambda):=\mathbb{E}[e^{-\lambda X}]$ for $\lambda>0$ and $\mu=0 $

What you did wrong is that $$ e^{-\lambda x}e^{-x^2/\sigma^2}\neq e^{\lambda/\sigma^2}e^{x^3} $$ does not hold (it seems that you that $e^ae^b=e^{ab}$ instead of $e^{a+b}$).
1 vote
Accepted

Why is the volume of the solid of revolution between two graphs $\int_{0}^{b}\pi[f^2(x)-g^2(x)]dx$ instead of $\int_{0}^{b}\pi[f(x)-g(x)]^2dx$?

For some intuition, consider revolving the region between the graphs of $y=y_0$ and $y=y_1$ on $[a,b]$. For definiteness let's say $y_1>y_0>0$. Now the region in space that you are creating is ...
  • 95.8k
0 votes

Why must the area under a curve require a non-negative function?

This is a purely pedagogical gambit. The definition is phrased like that precisely to stop people like you asking awkward questions! Later, presumably, your course will introduce the idea that the ...
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0 votes

Why must the area under a curve require a non-negative function?

Usually, a geometrical area is always between two curves or more. Let us consider the example of two curves $y=f(x)$ and $y=g(x)$. Further, if $f(x)> g(x)$ or $f(x) < g(x)$ in $(a,c)$; at $x=a,...
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1 vote

If $f:\mathbb{R} \to \mathbb{C}$ is a Borel function such that $|f|=1,$ then there exist $\alpha<\beta$ such that $\int_{\alpha}^{\beta}f(x)dx \neq 0$

Suppose not. Then for each bounded interval $I$, $f$ is integrable on $I$ and $\int_I f=0$. For a fixed $R$ and each finite union of disjoint open intervals $(I_k)_{k=1}^K$ contained in $(-R,R)$, $\...
1 vote
Accepted

Prove $\int_{0}^{+\infty} \frac{\exp(-t)}{\sqrt{\pi t}} \exp (-\frac{x^2}{4t}) ~\mathrm{d} t = e^{-|x|}$

Here's an approach that uses (mostly) basic calculus. Lemma (Glasser's master theorem): If $f$ is continuous, $\int_{-\infty}^{\infty} f$ exists, and $a\ge 0$ then $$\int_{-\infty}^{\infty} f\left(x-\...
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3 votes
Accepted

Solving the false "theorem" $\int\frac1{f(x)}dx=\frac1{\int f(x)dx}$

$y=-(-y^2)^{-1/2}\cdot\frac{dy}{dx}$ Here you forgot the term $-2f(x)$ from the previous equation. In general if you write $\int f(x)dx=\sqrt{-(f(x))^2}$ then you are already working with complex ...
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