I'm not exactly sure how to get started computing the limit of the improper Riemann integral

$$\lim_{\epsilon \rightarrow 0} \int_0^\infty \frac{\sin x}{x} \arctan\left(\frac{x}{\epsilon}\right)dx.$$

Using the result that $\int_0^\infty \frac{\sin x}{x} dx = \pi/2$, is there a way to interchange the limit and the integral to get $\pi^2/4$?

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    $\begingroup$ dominated convergenze? $\endgroup$ – john Jul 18 '15 at 15:41
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    $\begingroup$ The integral is not absolutely convergent. $\endgroup$ – Philip Hoskins Jul 18 '15 at 15:52
  • $\begingroup$ Numerically, the limit seems to be $\frac {\pi^2}4$. I shall post something tomorrow morning. $\endgroup$ – Claude Leibovici Jul 18 '15 at 17:41

By the dominated convergence theorem $$\lim_{\epsilon \to 0} \int_0^\pi \frac{\sin x}{x} \arctan\frac{x}{\epsilon}\,dx=\frac{\pi}{2}\int_0^\pi \frac{\sin x}{x}\,dx. $$ Now $$ \int_\pi^\infty \frac{\sin x}{x}\,\arctan\frac{x}{\epsilon}\,dx=\frac{\pi}{2}\int_\pi^\infty \frac{\sin x}{x}\,dx+\int_\pi^\infty \frac{\sin x}{x}\Bigl(\arctan\frac{x}{\epsilon}-\frac{\pi}{2}\Bigr)\,dx. $$ Let's stimate the second integral: $$ \Bigl|\arctan\frac{x}{\epsilon}-\frac{\pi}{2}\Bigr|=\int_{x/\epsilon}^\infty\frac{dt}{1+t^2}\le\frac{\epsilon}{x}, $$ and $$ \int_\pi^\infty \Bigl|\frac{\sin x}{x}\Bigl(\arctan\frac{x}{\epsilon}-\frac{\pi}{2}\Bigr)\Bigr|\,dx\le\epsilon\int_\pi^\infty\frac{|\sin x|}{x^2}\,dx. $$

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    $\begingroup$ That last integral does not converge, so I'm not sure what your result says. $\endgroup$ – Ron Gordon Jul 18 '15 at 16:18
  • $\begingroup$ This needs a little fixing. First, is it even clear that the integrals converge? Best to look at $\int_0^b$, then let $b\to \infty.$ Second, it's easy to see the integral from $0$ to $1$ does what it should. For the integral from $1$ to $b$, do what you did. $\endgroup$ – zhw. Jul 18 '15 at 16:23
  • $\begingroup$ @RonGordon but the bound does work from 1 to $\infty$ as well, so I think this is morally correct $\endgroup$ – Philip Hoskins Jul 18 '15 at 16:29
  • $\begingroup$ @RonGordon Thank you for the observation. I have fixed the argument. $\endgroup$ – Julián Aguirre Jul 18 '15 at 16:30
  • $\begingroup$ Thanks @JuliánAguirre. I fixed a typo for you in the first line. $\endgroup$ – Philip Hoskins Jul 18 '15 at 16:36

This is not an answer but it is too long for a comment.

I numerically computed the value of $$I_k= \int_0^\infty \frac{\sin x}{x} \arctan\left(\frac{x}{10^{-k}}\right)\,dx$$ and obtained the following results $$I_1=2.0946478630779095038$$ $$I_2=2.4071212109147673112$$ $$I_3=2.4590705601861920202$$ $$I_4=2.4663377878010322013$$ $$I_5=2.4672717431743382398$$ $$I_6=2.4673858619774465911$$ $$I_7=2.4673993461843410490$$ $$I_8=2.4674009018376888642$$ $$I_9=2.4674010781262894827$$ $$I_{10}=2.4674010978274761282$$ $$I_{20}=2.4674011002723396542$$ $$I_{30}=2.4674011002723396547$$ The limit is almost clearly $\frac{\pi^2}4$.

For $10\leq k\leq 100$, the plot of $\log(\frac{\pi^2}4-I_k)$ visually reveals an almost linear dependency to $k$ ($R^2=0.99999865$).

Later, to my surprize, a CAS found an explicit expression for the integral $$J(\epsilon)=\int_0^\infty \frac{\sin x}{x} \arctan\left(\frac{x}{\epsilon}\right)dx= \text{sgn}(\epsilon ) \,\frac{\sqrt{\pi }}{4} \,G_{2,4}^{3,1}\left(\frac{\epsilon ^2}{4}| \begin{array}{c} \frac{1}{2},1 \\ 0,\frac{1}{2},\frac{1}{2},0 \end{array} \right)$$ where appears the Meijer G function.

For $\epsilon >0$, a limited expansion of the above result gives $$J(\epsilon)=\frac{\pi ^2}{4}+\epsilon (\log (\epsilon )+\gamma -2)+\frac{1}{108} \epsilon ^3 (6 \log (\epsilon )+6 \gamma -13)+O\left(\epsilon ^4\right)$$ which almost perfectly reproduces the numerical results even for very small values of $k$.

  • $\begingroup$ Interesting! So does the $\log(\epsilon)$ in the first term say that the convergence is slightly worse than linear? That still seems pretty fast. $\endgroup$ – Philip Hoskins Jul 20 '15 at 2:00
  • $\begingroup$ I should agree if $\epsilon$ was not very small. $\endgroup$ – Claude Leibovici Jul 20 '15 at 5:03
  • $\begingroup$ Oh of course you're right. So it's better. Also, after reading up on the G function, it appears to be quite a beast :) $\endgroup$ – Philip Hoskins Jul 20 '15 at 5:12
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    $\begingroup$ @PhilipHoskins. I am sorry to disagree with you : it is not quite a beast. It is a MONSTER $\endgroup$ – Claude Leibovici Jul 20 '15 at 5:17

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