Evaluate $\int_0^\infty\frac{\sin^4x}{x^2}dx$ with $\int_0^\infty\frac{\sin x}{x}dx=\frac{\pi}{2}$$$$$ I use $\int_0^\infty\int_0^\infty e^{-x^2y}\sin x dy dx=\int_0^\infty\int_0^\infty e^{-x^2y}\sin x dx dy $ to evaluate the integration, I wonder how to use $\int_0^\infty\frac{\sin x}{x}dx=\frac{\pi}{2}$ to evaluate $\int_0^\infty\frac{\sin^4x}{x^2}dx$ more efficiently.
111619
I need help evaluating the following integral $$\int_{0}^{\infty} \frac{\sin^4 x}{x^2}\,dx$$ which should probably be equal to $\frac{\pi}{4}$  Using some trigonometric manipulations I got $\frac{3}{8} - \frac{\cos{2x}}{2} + \frac{\cos{4x}}{8}$ which using integration by parts doesn't lead me to anything pretty.  Update: Not sure if I should post this as a separate question but getting explanation why $\int_{0}^{\infty} \frac{\sin{ax}}{x} = \frac{\pi}{2}$ for a positive integer $a$ could help me solve this question.
123576
Given $$\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$$  but I can't find any clue then.
120543
Show:       $$ \lim_{n\to \infty} \int_{0}^{\frac{\pi }{2}} \sin^n(x) = 0. $$   I showed this using the . I want other methods to prove it. Thank you.
71393
Find a general formula for evaluating integrals of the form $\displaystyle\int_0^\infty x^a e^{-bx}dx.$   I've been pondering about this question for some time. Obviously integration by parts is way too inefficient. I know there's recursion involved and maybe there's a way to find a general formula through intuition.
46932
This refers back to $\int_{-\infty}^\infty \frac{\sin x}{x} \mathrm{d}x = \frac\pi2$ already posted. How do I arrive at $\frac\pi2$ using the residue theorem? I'm at the following point: $$\int \frac{e^{iz}}{z} - \int \frac{e^{iz}}{z},$$ and I would appreciate any help.
19082
Compute the integral.   $$\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})}$$
88402
I am trying to evaluate $$\int_{-\infty}^{\infty} \frac{\sin(x)^2}{x^2} dx $$ Would a contour work? I have tried using a contour but had no success. Thanks.  Edit: About 5 minutes after posting this question I suddenly realised how to solve it. Therefore, sorry about that. But thanks for all the answers anyways.
43485
Calculate the integral$$ \int_0^\infty \frac{x \sin rx }{a^2+x^2} dx=\frac{1}{2}\int_{-\infty}^\infty \frac{x \sin rx }{a^2+x^2} dx,\quad a,r \in \mathbb{R}. $$  Edit: I was able to solve the integral using complex analysis, and now I want to try and solve it using only real analysis techniques.
31614
How can I evaluate $$I=\int_{0}^{+\infty}\!e^{-ax^2-\frac b{x^2}}\,dx$$ for $a,b&gt;0$?  My methods:  Let $a,b &gt; 0$ and let $$I(b)=\int_{0}^{+\infty}e^{-ax^2-\frac b{x^2}}\,dx.$$ Then $$I'(b)=\int_{0}^{\infty}-\frac{1}{x^2}e^{-ax^2-\frac b{x^2}}\,dx.$$  What the other methods that can I use to evaluate it? Thank you.
78258
$$\int_{-\infty}^{\infty}\frac{x\sin x}{(x^2+a^2)(x^2+b^2)}\,\mathrm{d}x$$  This is easy to evaluate with complex analysis but is there an elementary way (substitution, partial fractions, integration by parts)?
111686
How to evaluate  $$  \int_0^\infty \frac{\sin(\varphi_1x)}{x}\frac{\sin\varphi_2x}{x} \cdots \frac{\sin\varphi_nx}{x} \frac{\sin(ax)}{x}\cos(a_1x) \cdots \cos(a_mx) \, dx  \text{ ?} $$  For small $n$ and $m$ it's simple (setting $\sin(kx)=\dfrac{e^{ikx}-e^{-ikx}}{2i}$ and using Jordan's lemma, but for arbitrary $n$ the calculation is too tedious. Surely there must be some nice trick here? Thanks
3970
I am looking for a short proof that $$\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$$ is also $\frac{\pi}{2}.$ Many proofs of this latter one are already in this .
88397
Evaluate integral $$\int_{0}^{1}{x^{-x}(1-x)^{x-1}\sin{\pi x}dx}$$ Well,I think we have $$\int_{0}^{1}{x^{-x}(1-x)^{x-1}\sin{\pi x}dx}=\frac{\pi}{e}$$  and  $$\int_{0}^{1}{x^{x}(1-x)^{1-x}\sin{\pi x}dx}=\frac{e\pi}{24}$$   With such nice result of these integral,why isn't worth to evaluate it?  I found a solution about the second one,but I wonder it will work for the first one   Note $$ S=\int_{0}^{1}{\sin{\pi x}x^{x}(1-x)^{1-x}dx}-\int_{0}^{1}{(1-x)e^{(i\pi+\ln{x}-\ln{(1-x)})x}dx} $$ Let $t=\ln{x}-\ln{(1-x)}$,$x=\frac{e^{t}}{1+e^{t}}$ Thus \begin{align} S&amp;=\int_{-\infty}^{+\infty}{\frac{1}{e^{t}+1}e^{(i\pi+t)\frac{e^{t}}{1+e^t}}\frac{e^{t}}{(1+e^{t})^{2}}dt}\\   &amp;=\int_{-\infty+i\pi}^{-\infty-i\pi}{e^{\frac{te^{t}}{e^{t}-1} } \frac{e^{t}}{(e^{t}-1)^{3}}dt} \end{align} Due to $$ f(z)=e^{\frac{te^{t}}{e^{t}-1} } \frac{e^{t}}{(e^{t}-1)^{3}},\qquad D=\{Z\in C|-\pi\leq Im(z) \leq \pi\}$$ Therefore $res(f,0)=-\frac{e}{24}$when $z=0$ with $ \zeta_{R}=\gamma_{R}+o_{R}+\tau_{R}$ $$\oint_{\zeta_{R}}{f(z)dz}=-2\pi i\cdot res(f,0)=\frac{2i\pi e}{24}$$ because $$ \{z_{n}\}\subset D,\qquad |z_{n}|\rightarrow\infty $$ Therefore $$ 2S=2\lim_{R\rightarrow \infty}\int_{\gamma_{R}}{f(z)dz} $$ gives $$ \int_{0}^{1}{\sin{\pi x}x^{x}(1-x)^{1-x}dx}=Im(S)=\frac{e\pi}{24} $$  My friend tian_275461 told me he use a simliar method to deal with the first one to obtain the result $\frac{\pi}{e}$,but I am not figure it out.
97606
Let $0&lt;\alpha&lt;2.$ Looking for a proof for the following: $$\int_0^{+\infty}\frac{\sin(x)}{x^\alpha}dx=\frac{\Gamma(\alpha/2)\Gamma(1-\alpha/2)}{2\Gamma(\alpha)}.$$ Any ideas?
52543
I have no idea how to start, it looks like integration by parts won't work.  $$\int_0^{\infty}\frac{x^5\sin(x)}{(1+x^2)^3}dx$$  If someone could shed some light on this I'd be very thankful.
53738
Here is my question  $$\int_0^{\infty} \frac{\log(x+\frac{1}{x})}{1+x^2}dx$$  I have tried it by substituting $x$ = $\frac{1}{t}$. I got the answer $0$ but the correct answer is $\pi log(2)$. Any suggestion would be appreciated.
123246
How can I calculate $$\int_0^\infty\frac{\sin x}xdx$$ by integration of a suitable function along the following paths:  where $R$ and $\varepsilon$ are the radius of the shown outer and inner semicircle, respectively. This seems to be quite interesting. How would one do that?