$$ I=\displaystyle \int_{0}^{\infty} \dfrac{\sin (\pi x^{2})}{\sinh^{2} (\pi x)} ~\mathrm{d}x $$ Found this on the cover of a book called "Integral Kokeboken" written in some language that i've never seen and btw the ans is  $$ I=\dfrac{2-\sqrt{2}}{4}$$
120781
I ran across this integral I get no where with. Can someone suggest a method of attack?.  $$\int_0^{\infty}\frac{\sin(\pi x^2)}{\sinh^2 (\pi x)}\mathrm dx=\frac{2-\sqrt{2}}{4}$$  I tried series, imaginary parts, and so forth, but have made no progress.    Thanks very much.
116992
During my studying to integration I find this integration.  So I tried to prove but I got stuk.  So I need help in this integration.  $$\displaystyle\int_{-\infty}^{\infty} \frac{x \sin (x)}{1+x^2}  dx = \frac {\pi}{e} $$
74829
I have seen $$\int_0^\infty \frac{\cos(x)}{x^2+1} \, dx=\frac{\pi}{2e}$$ evaluated in various ways.   It's rather popular when studying CA.   But, what about $$\int_0^\infty \frac{\sin(x)}{x^2+1} \, dx\,\,?$$  This appears to be trickier and more challenging.  I found that it has a closed form of  $$\cosh(1)\operatorname{Shi}(1)-\sinh(1)\text{Chi(1)}\,\,,\,\operatorname{Shi}(1)=\int_0^1 \frac{\sinh(x)}{x}dx\,\,,\,\, \text{Chi(1)}=\gamma+\int_0^1 \frac{\cosh(x)-1}{x} \, dx$$  which are the hyperbolic sine and cosine integrals, respectively.  It's an odd function, so  $$\int_{-\infty}^\infty \frac{\sin(x)}{x^2+1} \, dx=0$$  But, does anyone know how the former case can be done? Thanks a bunch.
52350
The following question comes from   post $$\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x$$ but now I'd be curious to know how to deal with it by methods of complex analysis. Some suggestions, hints? Thanks!!!  Sis.
20193
Find $\displaystyle \int_{0}^{\infty} \dfrac{\ln(x)}{1+x^2}dx$.   How should I change the limits of integration to evaluate this?
5682
show that  $$\int_{0}^{\infty} \frac {\sin^3(x)}{x^3}dx=\frac{3\pi}{8}$$  using different ways  thanks for all
69861
$$\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx$$   This was a Calc 2 problem for extra credit (we have done hyperbolic trig functions too, if that helps) and I didn't get it (don't think anyone did) -- how would you go about it?
68815
I have to evaluate:  $$\int_{0}^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\, \mathrm{d}x. $$ I can't get the right answer! So please help me out!
120467
$$problem:\int_0^\infty cos(x^2)\ln(x)dx$$ $$Re\int_0^\infty e^{-ix^2}ln(x)dx$$ $$\Gamma(s)=\int_0^\infty x^{s-1}e^dx, \ \ \ \ x=u^2, dx=2udu$$ $$2\int_0^\infty u^{2s-1}e^{-u^2}du$$ $$\Gamma'(s)=4\int_0^\infty u^{2s-1}e^{-u^2}ln(u)du$$ $$\frac{\Gamma'(1/2)}{4}=\int_0^\infty e^{-u^2}ln(u)du$$ $$Re\int_0^\infty e^{-ix^2}ln(x)dx$$ $$Re\frac{1}{\sqrt{i}}\int e^{-u^2}ln(u)-\frac{1}{\sqrt{i}}\int e^{-u^3}ln(\sqrt{i})du$$ $$Re[\frac{\Gamma'(\frac12)}{4\sqrt{i}}-\frac{\sqrt{\pi}ln(i)}{4\sqrt{i}}]=Re[\frac{\Gamma'\frac(12)-\sqrt{pi}\ln(i)}{4\sqrt{i}}]$$ $$Re[\frac{\Gamma(\frac12)\psi(\frac12)-\sqrt{\pi}\ln(i)}{4\sqrt{i}}]$$ $$Re[\frac{\sqrt{pi}(-\gamma-2\ln(2))-\frac{\pi^{\frac32}}{2}i}{4\sqrt{i}}]$$ $$=\frac{\sqrt{2\pi}}{8}(-\gamma-2\ln(2))-\frac{\pi^{\frac32}}{16}$$
26331
show that $$I=\int_{0}^{\sqrt{2\pi}}\sin{x^2}dx&gt;0$$  follow is my methods: let $$x^2=t$$ then $$I=\int_{0}^{\pi}\dfrac{\sin{t}}{2\sqrt{t}}dt+\int_{\pi}^{2\pi}\dfrac{\sin{t}}{2\sqrt{t}}dt=\int_{0}^{\pi}\dfrac{\sin{t}}{2}\left(\dfrac{1}{\sqrt{t}}-\dfrac{1}{\sqrt{\pi+t}}\right)dt$$ since $$\dfrac{1}{\sqrt{t}}-\dfrac{1}{\sqrt{\pi+t}}&gt;0$$ so $$I&gt;0$$  My quuestion: have other methods? Thank you
82074
How do you find the value of this integral, using real methods?  $$I=\displaystyle\int_{0}^{\pi/2} \log ({1+\cos x}) dx$$  The answer is $2C-\dfrac{\pi}{2}\log {2}$ where $C$ is Catalan's constant.
48343
The following integral has been on my mind for a while $$\int_0^\infty \frac{\sin(x)}{e^x-1}\,\mathrm d x \tag{$\dagger$}$$    Let us indicate the integrand as $f(x)=\frac{\sin(x)}{e^x-1}$. The following are a couple of observations.   The integrand can be extended by continuity in $0$ since $$\lim_{x\to 0}f(x)=\lim_{x\to 0}\frac{\frac{\sin(x)}{x}}{\frac{e^x-1}{x}}=1$$ This is the original reason I started playing around with this integral. Mathematica yields the result $$\int_0^\infty \frac{\sin(x)}{e^x-1}\,\mathrm d x=\frac{\pi}{2}\textrm{Coth}(\pi)-\frac 12 \approx 1.076674047$$ which, following numerical evidence, seems correct. Complex analysis may be useful here, since the integrand is a holomorphic function on $\mathbb C$. I tried writing $\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$, but I was not able to find an appropriate integration contour to solve the problem. The $-1$ in the denominator breaks the simmetry of the expression. This made most of my substitutions useless.   Can this integral be evaluated correctly, preferrably through complex analytic methods?
98316
I am trying to show  $$\int_{0}^{\infty}\frac{\sin^{3}(x)}{x^{3}}\mathrm dx = \frac{3\pi}{8}.$$  I believe the contour I should use is a semicircle in the upper half plane with a slight bump at the origin, so I miss the singularity.  Lastly I have the hint to consider  $$\int_{0}^{\infty}\frac{e^{3iz}-3e^{iz}+2}{z^{3}}\mathrm dz$$  around the contour I mentioned. Thanks for any help or hints!
3970
I am looking for a short proof that $$\int_0^\infty \left(\frac{\sin x}{x}\right)^2 \mathrm dx=\frac{\pi}{2}.$$ What do you think?  It is kind of amazing that $$\int_0^\infty \frac{\sin x}{x} \mathrm dx$$ is also $\frac{\pi}{2}.$ Many proofs of this latter one are already in this .
119970
$\displaystyle\int_0^0\frac 1x\:dx$ This integral is defined ? If it define, what is the value ?  Please,step by step
19082
Compute the integral.   $$\int_{-\infty}^\infty \frac{x^4}{1+x^8} \, dx$$  The answer at the back of the book is  $$\frac{\pi}{4\sin(\frac{3\pi}{8})}$$
45819
$$\int_{0}^{\infty}\frac{x\sin 2x}{9+x^{2}} \, dx$$   Some rearranging eventually gives $$\int_{0}^{\infty}\frac{x\sin 2x}{9+x^{2}} \, dx = \frac{-i}{2}\int_{-\infty}^{\infty} \frac{xe^{2ix}}{9+x^{2}}$$ Consider $f(z) = \frac{ze^{2iz}}{9+z^{2}}$ and the contour $\gamma$ of the semicircle laying in the upper half of the plane:   Let $\gamma_{R}$ denote the circular part with radius $R$ and $\gamma_{L}$ denote the part lying on the real axis with length $2R$.   Computing the residue at the only pole, $z = 3i$, we have that $$\oint_{\gamma}f(z) \, dz = \frac{i\pi}{e^{6}} $$ On the other hand, \begin{align*} \oint_{\gamma}f(z) \, dz &amp;= \oint_{\gamma_{R}}f(z) \, dz + \oint_{\gamma_{L}} f(z) \, dz \\ &amp;= \oint_{\gamma_{R}} f(z) \, dz + \int_{-R}^{R} \frac{xe^{2ix}}{9+x^{2}} \, dx \end{align*} We may evaluate the first integral in the usual way by parameterizing the contour and taking $z = Re^{i\theta}$. \begin{align*} \oint_{\gamma_{R}} f(z) \, dz = \int_{0}^{\pi} \, \frac{R^{2}e^{2i\theta}e^{2i\cos\theta}e^{-2R\sin\theta}}{(9+R^{2}e^{2i\theta})} d\theta \end{align*}  I'm aiming to show that this integral goes to $0$ as $R$ goes to infinity. This gives the desired result as the factor of $\frac{-i}{2}$ is all that is missing according to WolframAlpha. I'm not sure how to finish it though.