I have the integral:      $$\int_{-\infty}^\infty e^{-x^2} \, dx$$   And I’d like to solve it using differentiation under the integral sign. I understand that I must convert $e^{-x^2}$ to $e^{-x^2}g(x,t)$, where $g(x,t)$ is just some term involving a new parameter $t$. I tried multiplying by $e^{-2xt}$, which doesn’t really seem very logical, however, I just do not know what to put. How do I find the appropriate term to make the whole process work?
103316
I stumbled upon  on last weekend, it introduces an integral trick that exploits . On its last page, the author, Mr. Anonymous, left several exercises without any hints, one of them is to evaluate the Gaussian integral $$ \int^\infty_0 e^{-x^2} \,dx= \frac{\sqrt{\pi}}{2} $$ using this parametrization trick. I had been evaluating it through trial and error using different paramatrizations, but no luck so far.     Here are what I have tried so far:   A first instinct would be do something like:$$ I(b) = \int^\infty_0 e^{-f(b)x^2}\,dx $$ for some permissible function $f(\cdot)$, differentiating it will lead to a simple solvable ode: $$ \frac{I'(b)}{I(b)} = -\frac{f'(b)}{2f(b)} $$ which gives: $$ I(b) = \frac{C}{\sqrt{f(b)}}. $$ However, finding this constant $C$ basically is equivalent to evaluating the original integral, we are stuck here without leaving this parametrization trick framework. A second try involves an exercise on the same page: $$ I(b) = \int^\infty_0 e^{-\frac{b^2}{x^2}-x^2}dx. $$ Taking derivative and rescaling the integral using change of variable we have: $$ I'(b) = -2I(b). $$ This gives us another impossible to solve constant $C$ in: $$ I(b) = C e^{-2b} $$ without leaving this framework yet again. The third try is trying modify Américo Tavares's answer in this : $$ I(b) = \int^\infty_0 be^{-b^2x^2}\,dx. $$ It is easy to show that: $$ I'(b) = \int^\infty_0 e^{-b^2x^2}\,dx - \int^\infty_0 2b^2 x^2 e^{-b^2x^2}\,dx = 0 $$ by an integration by parts identity: $$ \int^\infty_0 x^2 e^{- c x^2}\,dx = \frac{1}{2c}\int^\infty_0 e^{- c x^2}\,dx . $$ Then $I(b) = C$, ouch, stuck again at this constant.     Notice in that  question, Bryan Yocks's answer is somewhat similar to the idea of parametrization, however he has to introduce another parametric integration to produce a definite integral leading to $\arctan$.  Is there such a one shot parametrization trick solution like the author Anonymous claimed to be "creative parameterizations and a dose of differentiation under the integral"?
45585
I'm self studying complex analysis. I've encountered the following integral:  $$\int_{- \infty}^{+ \infty} \frac{e^{ax}}{1+e^x} dx \text{ with } a \in \mathbb{R},\  0 \lt a \lt 1. $$  I've done the substitution $e^x = y$. What kind of contour can I use in this case ?
56394
I do not know how to solve this integral and then expand the result in powers of $t$  $$(1-t)^{-1/2}\int_{-\infty}^{\infty}\frac{\exp\left(\frac{2x^2t}{1+t}\right)}{1+x^2}dx=?,~~~~|t|&lt;1.$$  any idea?
83094
I am having trouble understanding where the following integral form comes from: $$\int_{-\infty}^{\infty} e^{-a x^2 }e^{-bx}=\sqrt{\frac{\pi}{a}}e^{\frac{b^2}{a}}$$ I see and understand that the value of the improper Gaussian integral is there, and I surmise that a change of variables is necessary when integrating, but I cannot figure out exactly how to prove this answer.
46239
This integral popped up recently  $$\int_{-\infty}^{\infty} \frac{e^{2x}-e^x}{x (1+e^{2x})(1+e^{x})}\mathrm{d}x = \log 2$$  A solution using both real and complex analysis is welcome. I tried rewriting it using symmetry, and then the series expansion of $1/(1+e^{nx})$, however this did not quite make it.   \begin{align*} I &amp; = 2\int_{0}^{\infty} \frac{e^{2x}-e^x}{x e^{3x}(1+e^{-2x})(1+e^{-x})}\mathrm{d}x \\ &amp; = 2 \int_{0}^{\infty}\frac{e^{-2x} -e^{-x}}{x} \left(\sum_{n=0}^\infty  (-1)^n e^{-n x}\right)\left(\sum_{m=0}^\infty (-1)^me^{-2mx}\right) \mathrm{d}x  \end{align*} The first part reminds me of a Frullani integral (it evaluates to $\log 2$). However I am unsure if this is the correct path, any help would be appreciated. =)
56507
I would like to evaluate the following integral $(a&gt;0)$      $$\int_{-\infty}^{\infty}\frac{\exp\left(-a x^2\right)}{x^2+b^2}dx.$$   I've have tried integration by parts, putting $e^{-ax^2}=u$, but I come across with this integral  $$\int_{-\infty}^{\infty}\exp\left(-a x^2\right)\arctan\left(\frac{x}{b}\right),$$  and I don't know how to do it. Could you help me?
79510
I think this question would be with complicated methods in singularity or other kind of methods. I need to solve the integral $$\int_0^\infty\frac{\cos(ax)-\cos(bx)}{x}dx$$ Thanks a lot.
44339
I'm having trouble calculating the integral  $$\int_{- \infty}^\infty \frac{\sin x}{x+i}\,dx $$  using residue calculus. I've previously encountered expressions of the form   $$\int_{- \infty}^\infty f(x) \sin x \,dx $$  where you would consider $f(z)e^{iz}$ on an appropriate contour (half circle), do away with the part of the contour that wasn't on the real axis by letting the radius go to infinity, then recover the imaginary part of the answer to get back the sine. However here, I can't replace the sine with $e^{iz}$ in my complex function because $$\operatorname{Im} \frac{e^{ix}}{x+i} \neq \frac{\sin x}{x+i}\cdots$$  How to remedy this? I'm not sure if substituting $\sin x = \frac{1}{2i}(e^{ix}-e^{-ix})$ and solving two integrals is how this problem is meant to be solved, although I'm 99% sure it would work
92455
$$\int_{-\infty}^{\infty} \frac{\cos(αx)}{(x^2+1)(x^2+4)} \mathrm dx. $$  I am not sure how to solve this question. Can anyone help me to approach this problem. Thanks.
42823
I'm trying to evaluate the improper integral,  $$I(a)=\int_{0}^{\infty}\frac{\tan^{-1}{x}}{e^{ax}-1}\mathrm{d}x,~~~\text{where }a\in\mathbb{R}^+.$$  Does this integral have a simple closed form expression? And if so, how best to obtain it?    My attempt  My first idea was to integrate by parts using $f=\tan^{-1}{x}$ and $dg=\frac{\mathrm{d}x}{e^{ax}-1}$, with the hope that the resulting integral would be amenable to solution by differentiating under the integral sign. So I found the indefinite integral,  $$\int\frac{\tan^{-1}{x}}{e^{ax}-1}\mathrm{d}x=\tan^{-1}{x}\left(\frac{1}{a}\log{(1-e^{ax})}-x\right)-\int\left(\frac{\log{(1-e^{ax})}}{a(x^2+1)}-\frac{x}{x^2+1}\right)\mathrm{d}x,$$  but then I realized that this would result in an imaginary boundary term for the corresponding definite integral over $[0,\infty)$ since,  $$\begin{cases}\lim_{x\to\infty}\tan^{-1}{x}\left(\frac{1}{a}\log{(1-e^{ax})}-x\right)=i\frac{\pi^2}{2a},\\ \lim_{x\to0}\tan^{-1}{x}\left(\frac{1}{a}\log{(1-e^{ax})}-x\right)=0.\end{cases}$$  I want to avoid complex variables if at all possible, so I don't know if I want to continue down this route.  Can anyone offer any hints or suggestions?    CORRECTION: As Vladimir pointed out, the correct anti-derivative is actually,  $$\int\frac{\tan^{-1}{x}}{e^{ax}-1}\mathrm{d}x=\tan^{-1}{x}\left(\frac{1}{a}\log{(e^{ax}-1)}-x\right)-\int\left(\frac{\log{(e^{ax}-1)}}{a(x^2+1)}-\frac{x}{x^2+1}\right)\mathrm{d}x.$$  Then, since,  $$\begin{cases}\lim_{x\to\infty}\tan^{-1}{x}\left(\frac{1}{a}\log{(e^{ax}-1)}-x\right)=\frac{\pi}{2}\cdot0=0,\\ \lim_{x\to0}\tan^{-1}{x}\left(\frac{1}{a}\log{-e^{ax}-1)}-x\right)=0,\end{cases}$$  we have,  $$I(a)=-\int_{0}^{\infty}\left(\frac{\log{(e^{ax}-1)}}{a(x^2+1)}-\frac{x}{x^2+1}\right)\mathrm{d}x.$$
53688
I'm trying to solve this integral: $$\int_{-\infty}^\infty e^{-x^2-x}dx$$ WolframAlpha shows this to be approximately $2.27588$. I tried to solve this by integration by parts, but I just couldn't get there. I'd be glad if someone could show me how to do it.   I've included my attempt at the problem below:  Note $\int_{-\infty}^\infty e^{-x^2-x} = \int_{-\infty}^\infty e^{-x^2}e^{-x}dx$. Then we can integrate by parts. Let $u(x) = e^{-x^2}$. Then $u'(x) = -2xe^{-x}$. Let $v'(x) = e^{-x}$. Then $v(x) = -e^{-x}$.  Then $u(x)v'(x) = u(x)v(x) - \int v(x)u'(x)dx$, i.e. $$\int_{-\infty}^\infty e^{-x^2}e^{-x}dx = -e^{-x^2}e^{-x} - \int e^{-x}2xe^{-x}dx = -e^{-x^2-x} - 2\int e^{-2x}xdx$$  Then we integrate $\int e^{-2x}xdx$ by parts. We pick $u(x) = x$, $u'(x) = dx$, $v'(x) = e^{-2x}$, and $v(x) = \int e^{-2x} dx = \frac {-1}{2} e^{-2x}$ by u-substitution. Skipping some steps, it follows that  $$\int e^{-2x}xdx = -\frac{1}{4}(e^{-2x})(2x+1)$$ Then  $$ -e^{-x^2-x} - 2\int e^{-2x}xdx = -e^{-x^2-x} + \frac{1}{2}(e^{-2x})(2x+1)$$ Which, when evaluated numerically, doesn't yield the desired result. So there's a problem here. Maybe someone else knows how to do this.
122196
How $$\int_{-\infty}^{\infty}\frac{dx}{1+x^2}$$ exists?   It is difficult question to me. i have tried to evaluate by using fact that $$\int_{-\infty}^{\infty} f(x) \ dx =\int_{-\infty}^{0} f(x)\, dx + \int_{0}^{\infty} f(x)\, dx$$ but i have failed in this one.   any hints on this one?
78427
I'm trying to find the value of the integral $\int_{-4}^{-2}e^{-x}\,dx$ but I just couldn't solve it.  Actually I found in a  that $\int e^x\,dx=e^x+C$ so I concluded:  $$ \int e^{-x}\,dx=\int\frac{1}{e^x}\,dx=\ln|e^x|$$  $$\int\limits_{-4}^{-2}e^{-x}\,dx=\left(\ln|e^{-2}|\right)-\left(\ln|e^{-4}|\right)=-2+4=2$$  I know the solution is wrong, but how can I solve this integral or any another integral like this?
39298
I'd love your help with finding the sign of the following integral: $\int_{0}^{2 \pi}\frac{\sin x}{x} dx$,I know that computing it is impossible.  I tried to use integration by parts and maybe to learn about the sign of each part and conclude something but It didn't work for me.  Any suggestions?
105702
I know that $$\int^\infty_0\frac x{e^x-1}dx=\frac{\pi^2}{6}$$ For substituting $u=2$ into $$\zeta(u)\Gamma(u)=\int^\infty_0\frac{x^{u-1}}{e^x-1}dx$$  However, I suspect that there is an easier proof, maybe by the use of complex analysis. I haven't learnt the  yet. All I know is the above formula and $\zeta(2)=\frac{\pi^2}{6}$. Can we can use the above integral to find out some of the $\zeta$'s value?
107304
The problem is to show  $$\int_0^\infty \frac{e^{-x}-e^{-xt}}{x}dx = \ln(t),$$  for $t \gt 0$.  I'm pretty stuck. I thought about integration by parts and couldn't get anywhere with the integrand in its current form. I tried a substitution $u=e^{-x}$ and came to a new integral (hopefully after no mistakes)  $$ \int_0^1 \frac{u^{t-1}-1}{\log(u)}du, $$  but this doesn't seem to help either. I hope I could have a hint in the right direction... I really want to solve most of it by myself.  Thanks a lot!
111197
$$\int_{-\infty}^{\infty} \frac{\cos(mx) dx}{e^{-x}+e^x} = \frac{\pi}{e^{m\pi /2}+e^{-m\pi /2}}$$  I need to evaluate the above integral specifically using a rectangluar contour and at some point applying the residue theorem. I should be ok with the residue part, but I'm confused at to how to get this into the example form shown in the book:  $$I =\int_{-\infty}^{\infty} \frac{e^{ax}dx}{1+e^x}=\frac{\pi}{\sin(a\pi)} $$  I tried using the identity $\cos (mx) = \frac{e^{ix}+e^{-ix}}{2}$ but I end up with $\frac{1}{2} \int_{-\infty}^{\infty}\frac{e^{ix}(e^{imx}+e^{-imx})dx}{1+e^{2ix}}$, which almost resembles the example but not quite...