# Linked Questions

0answers
224 views

### Integral $\int_0^{\infty} \frac{e^{-x^2}}{a+b\cos{x}}dx$

Hello there I am trying to solve for $a > b$: $$I=\int_0^{\infty} \frac{e^{-x^2}}{a+b\cos{x}}dx$$ My thought was to expand into fourier series $$g(t)=\frac{1}{a+b\cos t}$$ Since g(t) has the ...
0answers
62 views

### Suitable contour for an integral ($\Gamma(1/2)$)

Consider the following integral $$\int_0^\infty \frac{e^{-x}}{\sqrt{x}}dx=\sqrt{\pi}$$ This can be evaluated using contour integration methods. A similar question was asked before (unfortunately I ...
3answers
121 views

5answers
138 views

### Integrating this complicated integral for statistics [duplicate]

I want to show that : $$\int_{-\infty}^{\infty} e^\frac{-u^2}{2} du = \sqrt{2\pi}$$ Is there an elementary way using the tools of Calculus II to do this type of integration? I have not studied ...
2answers
1k views

### 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?
1answer
311 views

### 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}\...
2answers
4k views

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 $a^2=\... 1answer 428 views ### Prove the Wallis formula form$\left(4^{\zeta{(0)}} \cdot e^{-\zeta'{(0)}}\right)^2=\frac{\pi}{2}$How would you prove the following Wallis formula form $$\left(4^{\zeta{(0)}} \cdot e^{-\zeta'{(0)}}\right)^2=\frac{\pi}{2}?$$ Thanks in advance! 1answer 201 views ### Value of an integral related to Stirling's formula Consider the following improper integral : $$I = \int_1^\infty \left(\{t\}-\frac{1}{2}\right)\frac{dt}{t}.$$ Comparing with Stirling's formula, we can see that$I = \ln(\sqrt{2\pi}) - 1$. Is there ... 2answers 228 views ### Which is the easier way to do integration by parts when there is an exponential term? I am trying to calculate the following integral, and I would like to know if there is a general rule where we set either$u(x)$equal to the exponential term or$v'(x)\$ equal to the exponential term. ...

15 30 50 per page