how to prove this formula $\int_{0}^{\infty}\frac{e^{-\frac{x^2}{2}-\frac{z^2}{2x^2}}}{x^2}dx=\sqrt{\frac{\pi }{2}}\frac{ze^{-|z|}}{|z|}$? i have tried residue theorem, but it seems not to work
123576
Given $$\int_0^{\infty}e^{-x^2}dx = \frac{\sqrt{\pi}}{2}$$ evaluate:   $$\int_0^{\infty}e^{-a^2x^2-\frac{b^2}{x^2}}dx. $$   I can find that  $$\left(ax+\frac{b}{x}\right)^2 = a^2x^2+2ab+\frac{b^2}{x^2}$$  therefore: $$\int_0^{\infty}e^{-a^2x^2-\frac{b^2}{x^2}}dx = e^{2ab}\int_0^{\infty}e^{-\left(ax+\frac{b}{x}\right)^2}dx$$  but I can't find any clue then.
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.
73906
As a part of an exercise I need to calculate      $$ \lim_{r\to0}\int_{\sigma_{r}}\frac{e^{iz}}{z}\, dz $$      Where  $$ \sigma_{r}:\,[0,\pi]\to\mathbb{C} $$      $$ \sigma_{r}(t)=re^{it} $$   I know how to calculate the integral on the full circle ($|z|=r$) using Cauchy integral formula: $$2\pi ie^{iz}|_{z=0}=2\pi i$$  I have checked if  $$ (\overline{\frac{e^{iz}}{z})}=\frac{e^{iz}}{z} $$  so that the integral I want to calculate is $$\frac{1}{2}\cdot2\pi i=\pi i$$ but I got that that the equality above does not hold.  Can someone please suggest a way of calculating this integral ?
117095
I tried finding the integral of $\int_{|z|=2}^{}\frac{1}{z^2+1}dz$ but not sure whether it is correct. $\gamma(t)=2e^{it},t\in[0,2\pi]$  $$\int_{|z|=2}^{}\frac{1}{z^2+1}dz=\int_{0}^{2\pi}\frac{2ie^{it}}{4e^{2it}+1}dt$$  I cannot seem to take it forward from here. Is it possible to use Cauchy's integral formula to get an answer? Thanks
54941
$$ \int_0^{\infty } \frac{\log (x)}{e^x+1} \, dx = -\frac{1}{2} \log ^2(2) $$  Anyone an idea on how to prove this?
67392
I'm trying to solve this expression when $b \ne 0$:   $$ \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \frac{e^{-{x^2}/{2}}}{1+e^{-(ax+b)}}\,\mathrm dx $$   Using the answers provided to  we can solve $$\frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \frac{e^{-{x^2}/{2}}}{1+e^{-x}}\,\mathrm dx=\frac12$$  However, when we change the power of exponential term in the denominator to $-(ax+b)$ from $x$, the expression can not be solved using the same methods used in the . I tried solving by substituting $z=ax+b$ and it also did not work.
12892
Proving that $$\sum_{n=0}^{\infty }\frac{1}{(2n)!!}=\sqrt{e}$$ Firstly, I tried to check the value with the exponential function at $x=.5$ but I found its terms not equal to the series terms.
113167
Let's consider the following integral $$\int \limits_{-\infty}^{\infty} \frac{\cos2x}{(x^2+4)^2} \, dx. \tag{1}$$  I would like to compute $(1)$ using residue theory. Let's consider a complex function $$f(z) = \frac{\cos2z}{(z^2+4)^2} = \frac{\cos2x}{(z-2i)^2(z+2i)^2}.$$  Of course $\text{Im}(-2i) &lt; 0$ thus I am to calculate residue only in the point $z_0 = 2i$.  Noting that $z_0$ is a double pole we have $$R = \text{res}_{z_0}f(z) = \lim_{z \to 2i} \frac{d}{dz} \bigg((z-2i)^2 \frac{\cos(2z)}{(z-2i)^2(z+2i)^2} \bigg).$$  After some calculations we get $$R = \frac{i \big(-5 + 3 e^8 \big)}{64 e^4}.$$ That implies $$\int \limits_{-\infty}^{\infty} \frac{\cos2x}{(x^2+4)^2} \, dx = 2 \pi i \frac{i \big(-5 + 3 e^8 \big)}{64 e^4} = -\frac{(-5 + 3 e^8) \pi}{32 e^4}.$$  According to  $(1)$ is equal to $$\frac{5 \pi}{16 e^4}.$$  What am I doing wrong?
114669
I am trying to solve:  $$\int_{-\infty}^{\infty}\frac{x\sin{x}}{x^2 + a^2}dx$$  and show that it equals $\pi e^{-a}$ for all $a &gt; 0$.  I translate it to its complex equivalent:  $$lim_{R \rightarrow \infty} \int_{-R}^{R}\frac{z\sin{z}}{z^2 + a^2}dz = \int_{C}\frac{z\sin{z}}{z^2 + a^2}dz - \int_{\gamma}\frac{z\sin{z}}{z^2 + a^2}dz$$  where C is a half circle with radius R, above the real axis, and $\gamma$ is only the curved (top part) of the half circle.  I've shown that $$\int_{\gamma}\frac{z\sin{z}}{z^2 + a^2}dz$$ tends to 0 as $R \rightarrow \infty$.  So now:  $$lim_{R \rightarrow \infty} \int_{-R}^{R}\frac{z\sin{z}}{z^2 + a^2}dz = \int_{C}\frac{z\sin{z}}{z^2 + a^2}dz$$  By residue theorem,   $$\int_{C}\frac{z\sin{z}}{z^2 + a^2}dz = 2\pi i \cdot res_{ai}f$$  as $ai$ is the pole of $f$ that lies above the real axis.  $$2\pi i \cdot res_{ai}f = 2\pi i \cdot lim_{z \rightarrow ai} (z-ai)\frac{z sin{z}}{z^2 + a^2}$$  By L'Hospital's rule, this is equivalent to:  $$2\pi i \cdot lim_{z \rightarrow ai} (z-ai)\frac{z^2\cos{z} + 2z\sin{z} - zai\cos{z} - ai\sin{z}}{2z} = 2\pi i \cdot \frac{\sin{ai}}{2} = \pi i \sin{ai}$$  But how is $\pi i \sin{ai}$ equivalent to $\pi e^{-a}$? Did I make a mistake?
45721
Now we want to prove: $$\sum_{n=-\infty}^{\infty}\frac{1}{(n+\alpha)^2}=\frac{\pi^2}{(\sin \pi \alpha)^2}$$ $\alpha$ >0 and not an integer. According to poisson summation formula $$\sum_{n=-\infty}^{\infty}f(x+n)=\sum_{n=-\infty}^{\infty}\hat{f}(n)e^{2\pi inx}.$$ So, if we let $f(x)=\frac{1}{x^2}$, then $$\sum_{n=-\infty}^{\infty}f(x+n)=\sum_{n=-\infty}^{\infty}\frac{1}{(n+x)^2}$$ Therefore, if we can prove that $$\sum_{n=-\infty}^{\infty}\left(\int_{-\infty}^{\infty}\frac{1}{x^2} e^{-2\pi inx}dx e^{2\pi inx}\right)=\frac{\pi^2}{(\sin \pi x)^2}$$ Then it will be done. But I don't know how to prove $$\sum_{n=-\infty}^{\infty}\left(\int_{-\infty}^{\infty}\frac{1}{x^2} e^{-2\pi inx}dx e^{2\pi inx}\right)=\frac{\pi^2}{(\sin \pi x)^2}$$ Who could give me some hints? Thanks!
12916
How do I compute      $$\int_0^\infty \frac{(\log x)^2}{1+x^2} dx$$   What I am doing is take   $$f(z)=\frac{(\log z)^2}{1+z^2}$$  and calculating   $\text{Res}(f,z=i) = \dfrac{d}{dz} \dfrac{(\log z)^2}{1+z^2}$ which came out to be $\dfrac{\pi}{2}-\dfrac{i\pi^2}{8}+\dfrac{i\pi}{2}$  Im not too sure how to move on from here. the given answer is $\dfrac{\pi^3}{8}$  Any help will be appreciated. thank you in advanced.
97412
I'm looking for a way to prove $$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{(-1)^m E_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}$$ I know that $$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{1}{4^{2m+1}}\left(\zeta\left(2m+1,\frac14\right)-\zeta\left(2m+1,\frac34\right)\right)$$ so maybe I could simplify the above more?
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...
24079
Find the value of $\displaystyle\int_{|z|=2}\frac{e^{1/z^2}}{1-z}dz$   The first idea I had was to apply Cauchy's integral formula. But can this be done? $e^{1/z^2}$ is not holomorphic on the interior of $|z|=2$. Still, if we would apply this we would obtain:  $\displaystyle\int_{|z|=2}\frac{e^{1/z^2}}{1-z}dz=-2\pi ie$.  Another way I think, is to apply the residue theorem. For $1-z$ has a simple zero at $z=1$ and $e^{1/1^2}\neq 0$, then the residue of $e^{1/z^2}/(1-z)$ at $z=0$ is $-e$. Again we obtain $2\pi iRes(g;0)=-2\pi ie$, where $g=\frac{e^{1/z^2}}{1-z}$.  But again, it seems to me this is not right because $g$ has two singularities, at $z=1$ and $z=0$. What we might do about $z=0$? The answer is not $-2\pi i e$?
17907
Most of us are aware of the classic Gaussian Integral   $$\int_0^\infty e^{-x^2}\, dx=\frac{\sqrt{\pi}}{2}$$  I would be interested in evaluating the similar sum  $$\sum_{x=0}^\infty e^{-x^2}$$  Now, because $\exp(-\lfloor x \rfloor^2) \ge \exp(-x)$, we find   $$\sum_{x=0}^\infty e^{-x^2}= \int_0^\infty e^{-\lfloor x \rfloor^2}\, dx \ge \int_0^\infty e^{-x^2}\, dx=\frac{\sqrt{\pi}}{2}$$  Does a closed form for this sum exist?  If so, what would it be?  I would be very interested in how a closed form would be found for this function.
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?
11415
Prove the following using complex analysis   $$\tag{1}\int^\infty_0\frac{\ln(1+x)}{1+x^2}\,dx=\frac{\pi}{2}\ln(2)$$  I found this problem in Schaum's outlines of complex variables.   I thought that we could solve the problem using a similar method to evaluating   $$\int^\infty_0\frac{\ln(1+x^2)}{1+x^2}\,dx$$   Where we integrate the function  $$f=\frac{\log(1-iz)}{z^2+1}$$  around half a circle in the upper half plane and chosing an analytic branch of the logarithm in the region.   I spent so much time on (1) without success.