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I was surprised by the value of an integral I saw in Wolfram Alpha so I wanted to prove it on my own. This is the following integral : $$ \displaystyle I=\int_{0}^{+\infty}\frac{\sin^2\left(t\right)}{1+t^2}\text{d}t=\frac{\pi}{2e}\text{sinh}\left(1\right)$$

Is there an easy way to show this equality ?

In order to prove it, my idea was to introduce its twin integral $ \displaystyle J=\int_{0}^{+\infty}\frac{\cos^2\left(t\right)}{1+t^2}\text{d}t$. Then we have two results : $$I+J=\frac{\pi}{2}$$ $$I-J=\int_{0}^{+\infty}\frac{\cos\left(2t\right)}{1+t^2}\text{d}t$$ I would like to show that for $w>0$ $$\int_{0}^{+\infty}\frac{\cos\left(wt\right)}{1+t^2}\text{d}t=\frac{\pi}{2}e^{-t}$$ I've searched for a differential equation with its derivative $\displaystyle -\int_{0}^{+\infty}\frac{t\sin\left(wt\right)}{1+t^2}\text{d}t$ but it doesnt seem that nice even when I integrate by parts... Any ideas ?

Atmos
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  • May I know what methods you know? What was wrong with the differential equation you got? You can see in this answer that it is not difficult at all https://math.stackexchange.com/a/2480371/349501 – Shashi Dec 26 '17 at 12:53
  • By the way besides that answer I linked, there are other beautiful answers in the same post using the Residue Theorem. – Shashi Dec 26 '17 at 12:56
  • Thanks for the answer. Maybe i was just blind huhu. But i wanted to avoid using Dirichlet's integral ( and residues ) ^^ – Atmos Dec 26 '17 at 12:57
  • I do not understand why you would like to avoid something that helps you so much, plus it is solvable with simple methods. So what exactly do you want? If you want other methods you can look it up. In some posts they use the Laplace Transform and fubini etc – Shashi Dec 26 '17 at 13:02
  • Do you have a link for the Lasplace Transform ? I mean it's to make a proof suitable with a mathematics programe which does not even use contours or complex analysis. Furthermore, i would have to prove Dirichet's integral which means again Laplace Transform or limits etc ... I thought ( seeing the simple differential equation ) that it would be really easy to prove that f'=-f and not show that f''=f. – Atmos Dec 26 '17 at 13:10
  • I do not have it in "hand range" but maybe you can try it yourself. The Laplace Transform needs inversion, do these students know how to do that? Or use Laplace transform tables at least? – Shashi Dec 26 '17 at 13:14
  • They know how to swap limits and integrals as well as differentiate under integrals. – Atmos Dec 26 '17 at 13:17
  • Your first expression on the RHS is wrong btw you must devide the whole thing by $e$. – Shashi Dec 26 '17 at 14:53

1 Answers1

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Too long for a comment. Let $\omega\geq 0$ and define: \begin{align} f(\omega)= \int^\infty_0 \frac{\cos(\omega t)}{t^2+1}\,dt \end{align} The Laplce transform of it: \begin{align} \mathcal{L}(f)(s)=\int^\infty_0 \int^\infty_0 \frac{\cos(\omega t)}{t^2+1}e^{-s\omega}\,dt\,d\omega &= \int^\infty_0 \int^\infty_0 \frac{\cos(\omega t)}{t^2+1}e^{-s\omega}\,d\omega\,dt\\ &= \int^\infty_0 \frac{s}{(t^2+1)(s^2+t^2)}\,dt\\ &=\frac{s}{s^2-1}\int^\infty_0 \frac{1}{1+t^2}-\frac{1}{t^2+s^2}\,dt\\ &=\frac{s}{s^2-1} \left( \frac{\pi}{2} -\frac{\pi}{2s}\right)\\ &=\frac{\pi}{2}\left( \frac{s}{s^2-1}-\frac{1}{s^2-1}\right)\\ &=\frac{\pi}{2}\frac{1}{s+1} \end{align} assuming $s>0$. Now inverting that yields: \begin{align} f(\omega) = \frac{\pi}{2}e^{-\omega} \end{align} For this method you must know the Laplace Transfrom and inverting it (this one was relatively easy though). Finally notice that some steps need to be filled which I leave it to you.

Shashi
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