# Tag Info

## New answers tagged integration

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• 11.7k
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 ...
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• 685
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### 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 ...
• 1,125
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### 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 ...
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• 3,321
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• 8,377
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### 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 $$0 \leq \xi \leq L, \xi \leq x \leq L$$
• 422
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 ...
• 33.8k
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### 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 ...
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### Expected distance of a point from the center of a circle

$$\frac 1{\pi R^2}\int_{x^2+y^2<R^2}\sqrt{x^2+y^2}dx dy = (polar\ coordinates) = \frac 1{\pi R^2}\int_{r<R}2\pi r^2 dr=\frac 23 R$$
• 1,884
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### 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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• 7,419
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• 40.5k
1 vote

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 $(-R,R)$, $\... • 161k 1 vote Accepted ### Prove$\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-\... • 6,265 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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