# Linked Questions

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1answer
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### Evaluating $\int_0^{\infty}e^{-\alpha x^2 \cos \beta} \cos(\alpha x^2 \sin \beta) dx$

Q: Suppose $\alpha>0$ and $|\beta|<\pi/2$, show that \begin{align*} \textbf{(1)} \; \int_0^{\infty}e^{-\alpha x^2 \cos \beta} \cos(\alpha x^2 \sin \beta) dx &= \frac 1 2 \sqrt{\pi/\alpha}\...
5answers
655 views

### The other ways to calculate $\int_0^1\frac{\ln(1-x^2)}{x}dx$

Prove that $$\int_0^1\frac{\ln(1-x^2)}{x}dx=-\frac{\pi^2}{12}$$ without using series expansion. An easy way to calculate the above integral is using series expansion. Here is an example \begin{...
1answer
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2answers
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### How to compute the integral $\int_{-\infty}^\infty e^{-x^2}\,dx$? [duplicate]

How to compute the integral $\int_{-\infty}^\infty e^{-x^2}\,dx$ using polar coordinates?
2answers
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### Complex Analytic Proof of the Gaussian Integral $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$

Prove that $\int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}$. Here is my attempted solution: Define $a:=\sqrt{\pi}e^{\frac{\pi i}{4}}$ and let $f(z) = \frac{e^{-z^2}}{1+e^{-2az}}$. Note that \$a^2=\...

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