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Evaluate $$\int_{-\infty}^{\infty}\arctan\left(\frac{1}{2x^2}\right)\mathrm dx$$

And how can I solve it using $$\sum^{\infty}_{x=-\infty}\arctan\left(\frac{1}{2x^2}\right)\quad\text{ and }\quad \sum^{\infty}_{x=1}\arctan\left(\frac{1}{2x^2}\right)$$

My try:
Let $$ I = \int_{-\infty}^{\infty}\arctan\left(\frac{1}{2x^2}\right)\cdot 1 \ \mathrm dx = 2\int_{0}^{\infty}\arctan\left(\frac{1}{2x^2}\right)\cdot 1\ \mathrm dx $$ Using Integration by parts, we get $$I = 2\left[\arctan\left(\frac{1}{2x^2}\right)\cdot x\right]_{0}^{\infty}-2\int_{0}^{\infty}\frac{4x^2}{1+4x^4}\ \mathrm dx$$

Now how can I solve after that and also how we connect these integrals to the sums?

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  • $\begingroup$ Where is $r$ used in those sums? Did you mean $r$ instead of $x$ in those sums? $\endgroup$ – Thomas Andrews Aug 30 '15 at 12:31
  • $\begingroup$ try it by parts $\endgroup$ – Dr. Sonnhard Graubner Aug 30 '15 at 12:35
  • $\begingroup$ Why do you think the integral is related to the sums? $\endgroup$ – GEdgar Aug 30 '15 at 13:06
  • $\begingroup$ @GEdgar: it is a natural thought. The integral is just the "continuous version" of the series, so it is reasonable to expect that if we are able to compute one, we are able to compute the other, too. In general, things do not work so smooth, but in this particular case we are lucky to be right about it. $\endgroup$ – Jack D'Aurizio Aug 30 '15 at 13:16
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    $\begingroup$ I think it is unreasonable to expect a connection. Integral $\int_1^\infty (1/x^3)\,dx=1/2$ but I doubt that helps in evaluating $\sum_{n=1}^\infty 1/n^3 = \zeta(3)$. $\endgroup$ – GEdgar Aug 30 '15 at 13:19
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You do not need to connect the integral to the sum in any way, it is just the process in dealing with them that is the same. We are able to compute $$\sum_{n\geq 1}\arctan\frac{1}{2n^2}$$ since it is a telescopic series: $$\arctan\frac{1}{2n^2}=\arctan\frac{1}{2n-1}-\arctan\frac{1}{2n+1}$$ so we are applying summation by parts. What's the continuous version of summation by parts? Integration by parts, obviously. So the discrete case points out the right way for dealing with the continuous case. As you noticed, through integration by parts the last problem boils down to computing: $$ I=\int_{0}^{+\infty}\frac{x^2\,dx}{1+4x^4},$$ and since $$(1+4x^4)=(1+4x^2+4x^4)-(4x^2)=(1+2x^2)^2-(2x)^2=(1+2x+2x^2)(1-2x+2x^2),$$ it is not difficult to compute $I$ through partial fraction decomposition and/or the residue theorem. At last we have $I=\frac{\pi}{8}$.

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  • $\begingroup$ It should be noted that that factorization is far from obvious. (Until, of course, one sees it for the first time, in which case it makes one wonder how one didn't think of it before.) $\endgroup$ – Akiva Weinberger Aug 30 '15 at 13:41
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The similarity between the sum $\displaystyle \sum_{k=-\infty}^{\infty}\arctan\left(\dfrac{1}{2k^{2}}\right)$ and the integral $\displaystyle \int_{x=-\infty}^{\infty}\arctan\left(\dfrac{1}{2x^{2}}\right)\, dx$ can be explained by Poisson's summation formula $$ \sum_{k=-\infty}^{\infty}f(k) = \sum_{n=-\infty}^{\infty}\hat{f}(2\pi n) $$ where $f$ is a sufficiently summable and integrable function and $$ \hat{f}(\xi) = \int_{-\infty}^{\infty} e^{-i\xi x}f(x)\,dx \quad \text{(the Fourier transform)}. $$ We observe that $$ \sum_{n=-\infty}^{\infty}\hat{f}(2\pi n) = \hat{f}(0) + \sum_{n \neq 0}\hat{f}(2\pi n) = \int_{-\infty}^{\infty} f(x)\, dx + \sum_{n \neq 0}\hat{f}(2\pi n). $$ Consequently $$ \sum_{k=-\infty}^{\infty}f(k) = \int_{-\infty}^{\infty} f(x)\, dx $$ if and only if $$ \sum_{n \neq 0}\hat{f}(2\pi n) = 0. $$ In this case $\displaystyle f(x) = \arctan\left(\dfrac{1}{2x^{2}}\right)$ and $\displaystyle \hat{f}(\xi) = 2\pi\dfrac{\sin(|\xi|/2)}{|\xi|}\exp(-|\xi|/2)$. Apparently $\hat{f}(2\pi n)=0$ for all integers $n\neq 0$.

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