Linked Questions

1
vote
0answers
193 views

Residue calculus: $\int_{-\infty}^\infty e^{-x^2} \mathrm{d}x$ [duplicate]

I am pretty sure I have read the answer somewhere on this site, but sadly I am unable to find the question. How to evaluate $\int_{-\infty}^\infty e^{-x^2} \mathrm{d}x$ using the residue theorem?
27
votes
5answers
101k views

Integral: $\int_{-\infty}^{\infty} x^2 e^{-x^2}\mathrm dx$

I don't know how to evaluate it. I know there is one method using the gamma function. BUT I want to know the solution using a calculus method like polar coordinates. $$\int_{-\infty}^\infty x^2 e^{-x^...
8
votes
7answers
7k views

How to prove $ \int_0^\infty e^{-x^2} \; dx = \frac{\sqrt{\pi}}2$ without changing into polar coordinates?

How to prove $ \int_0^\infty e^{-x^2} \; dx = \frac{\sqrt{\pi}}2$ other than changing into polar coordinates? It is possible to prove it using infinite series?
15
votes
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{...
6
votes
6answers
3k views

How to integrate $\displaystyle 1-e^{-1/x^2}$?

How to integrate $\displaystyle 1-e^{-1/x^2}$ ? as hint is given: $\displaystyle\int_{\mathbb R}e^{-x^2/2}=\sqrt{2\pi}$ If i substitute $u=\dfrac{1}{x}$, it doesn't bring anything: $\,\...
1
vote
3answers
2k views

$\int e^{-x^2}dx$ [duplicate]

Possible Duplicate: Proving $\int_{0}^{\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2}$ How does one integrate $\int e^{-x^2}\,dx$? I read somewhere to use polar coordinates. How is this done? What ...
6
votes
2answers
4k views

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=\...
11
votes
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!
0
votes
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?
9
votes
1answer
1k views

Evaluating definite integrals

This question came up when I was reading through this question. Are there definite integrals which cannot be computed using any real analysis techniques but are amenable using only complex analysis ...
1
vote
1answer
593 views

Computing the integral of $e^{-x^2}$ over the entire line [duplicate]

Possible Duplicate: Proving $\\int_{0}^{+\\infty} e^{-x^2} dx = \\frac{\\sqrt \\pi}{2}$ At lunch with a math friend years ago, he showed me an integral whose solution was, he said, so beautiful ...
5
votes
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}\...
5
votes
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 ...
-2
votes
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 ...
2
votes
2answers
106 views

Strange integrand

Is it possible, and if yes, how, to evaluate an integral like $\int \sqrt{x} e^{x}dx$? I have hear of the Gaussian function which integrates to $\sqrt{\pi}$ but what about this? Thank you.
3
votes
1answer
211 views

$\sum_{n=-\infty}^\infty e^{-\alpha n^2+\beta n}$

Hi I am trying to calculate the sum given by $$ \sum_{n=-\infty}^\infty e^{-\alpha n^2+\beta n}=\ = \sqrt{\frac{\pi}{\alpha}} e^{\beta^2/(4\alpha)} \vartheta_3\big(-\frac{\pi\beta}{2\alpha},e^{-\pi^2/(...
3
votes
3answers
121 views

Improper integral with complex limits

I would like to compute an integral of the form ($a,b \neq 0$) $$\int_{-\infty}^{\infty} e^{-(ax+ib)^2} dx = \frac{1}{a} \int_{-\infty+ib}^{\infty+ib} e^{-z^2} dz$$ where we made the substitution $z ...
1
vote
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. ...
5
votes
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 ...
0
votes
1answer
49 views

Computing the integral $\int e^{2i(a-b)x}e^{-\frac{1}{2}x^2}\mathrm{d}x$

I am pretty stuck when I tried to calculate the Wigner function for the coherent state. Below is part of the equation that I find very challenging. $$ \int e^{2i(a-b)x}e^{-\frac{1}{2}x^2}\mathrm{d}x\...
1
vote
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 ...