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(I'm leaving the photo in case I mistyped something) enter image description here

$\int_0^\infty \frac{sin(x)}{x} dx$ $ = $ $\int_0^{2\pi}\sum_{n=0}^{\infty}\frac{sin(x)}{x+2n\pi}dx$ $ = $ $\int_{0}^{\pi}sin(x)(\sum_{n=0}^\infty\frac{1}{x+2n\pi}-\frac{1}{2\pi-x+2n\pi})dx$ $ = $
$\int_{0}^{\pi}sin(x)(\frac{1}{x}+\sum_{n=1}^{\infty}(\frac{1}{x+2n\pi}+\frac{1}{x-2n\pi}))dx$ $ = $ $\int_{0}^{\pi}\frac{sin(x)}{x}dx+\int_{0}^{\pi}2xsin(x)(\sum_ {n=1}^{\infty}\frac{1}{x^2-(2n\pi)^2})dx$

I'm trying to integrate $sin(x)/x$ from $0$ to $\infty$. I do that by first noticing that I can rewrite it as an integral from $0$ to $2 \pi$ and an infinite series that doesn't converge. My problem there is that I'm not sure how or even if I have to normalise the differential dx now that I'm using a shorter interval and an infinite divergent series. After that I simply do some algebraic stuff and I end up getting two different integrals. One is basically the same but only from $0$ to $\pi$ and that is where I thought that this method wasn't going to work unless I try one of the basic methods that already work with the original integral. What I thought was more interesting is how one could go about finding a different proof for the Basel problem that appears as part of the integral on the right.

BTW if you have problems with my handwriting please let me know.

David
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  • Don't take pictures of the math problem. Write it out using mathjax. https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – spinosarus123 Oct 10 '24 at 21:31

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From your result to the answer, you can use the Herglotz trick (see here):

\begin{align*} \pi\cot(\pi u) &= \sum_{n=-\infty}^\infty \frac1{u+n} \\ \frac{\cot(x/2)}{4x} &= \frac1{2x^2} + \sum_{n=1}^\infty \frac1{x^2-(2\pi n)^2}\\ \int_0^\infty \frac{\sin x}x dx &= \int_0^\pi \left[ \frac{\sin x}x + 2x\sin x \left[\frac{\cot(x/2)}{4x} - \frac1{2x^2}\right]\right\}\\ &= \frac12\int_0^\pi \sin x \cot(x/2) \\ &= \frac\pi2 \end{align*}

Ramashalanka
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  • Is there a problem with normalising the differential at the beginning? That's what's making me doubt the rest – David Oct 10 '24 at 22:42
  • I would start from your third line: think of a single term in the sum being the integral of the region $\pi$ to the right and $\pi$ to the left. – Ramashalanka Oct 10 '24 at 22:55
  • Yeah I see that already if I understood what you meant. Oh wait of course because of the distributive property every element of the sum is multiplied by the differential so there is no need for normalisation. – David Oct 10 '24 at 22:59
  • Just out of curiosity what would be a standard way of solving the Basel problem from something like this? – David Oct 10 '24 at 23:01
  • Standard way? I don't know: a new question? – Ramashalanka Oct 10 '24 at 23:06