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### 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?
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. $$\int_{-\infty}^\infty x^2 e^{-x^... 5answers 649 views ### The other ways to calculate \int_0^1\frac{\ln(1-x^2)}{x}dx Prove that$$\int_0^1\frac{\ln(1-x^2)}{x}dx=-\frac{\pi^2}{12}$$without using series expansion. An easy way to calculate the above integral is using series expansion. Here is an example \begin{... 7answers 6k views ### How to prove  \int_0^\infty e^{-x^2} \; dx = \frac{\sqrt{\pi}}2 without changing into polar coordinates? How to prove  \int_0^\infty e^{-x^2} \; dx = \frac{\sqrt{\pi}}2 other than changing into polar coordinates? It is possible to prove it using infinite series? 6answers 3k views ### How to integrate \displaystyle 1-e^{-1/x^2}? How to integrate \displaystyle 1-e^{-1/x^2} ? as hint is given: \displaystyle\int_{\mathbb R}e^{-x^2/2}=\sqrt{2\pi} If i substitute u=\dfrac{1}{x}, it doesn't bring anything: \,\... 3answers 2k views ### \int e^{-x^2}dx [duplicate] Possible Duplicate: Proving \int_{0}^{\infty} e^{-x^2} dx = \frac{\sqrt \pi}{2} How does one integrate \int e^{-x^2}\,dx? I read somewhere to use polar coordinates. How is this done? What ... 2answers 4k views ### Complex Analytic Proof of the Gaussian Integral \int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi} Prove that \int_{-\infty}^{\infty}e^{-z^2}dz=\sqrt{\pi}. Here is my attempted solution: Define a:=\sqrt{\pi}e^{\frac{\pi i}{4}} and let f(z) = \frac{e^{-z^2}}{1+e^{-2az}}. Note that a^2=\... 1answer 425 views ### Prove the Wallis formula form \left(4^{\zeta{(0)}} \cdot e^{-\zeta'{(0)}}\right)^2=\frac{\pi}{2} How would you prove the following Wallis formula form$$ \left(4^{\zeta{(0)}} \cdot e^{-\zeta'{(0)}}\right)^2=\frac{\pi}{2}?Thanks in advance! 2answers 977 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 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 ... 1answer 584 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 ... 1answer 310 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}\... 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}. $$Comparing with Stirling's formula, we can see that I = \ln(\sqrt{2\pi}) - 1. Is there ... 5answers 137 views ### Integrating this complicated integral for statistics [duplicate] I want to show that :$$ \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 ...
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.