Evaluation of $ \int_{-\infty}^{\infty}\arctan (\frac 1{2x^2})\ \mathrm dx$

Question

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?

Answer 1

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 $n\neq 0$.

Answer 2

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 $I=\frac{\pi}{8}$.

Answer 3

$$\begin{align*} & \int_{-\infty}^\infty \arctan\left(\frac1{2x^2}\right) \, dx \\ &= 2 \int_0^\infty \arctan\left(\frac1{2x^2}\right) \, dx \\ &= \sqrt2 \int_0^\infty \arctan\left(x^2\right) \, \frac{dx}{x^2} \tag1 \\ &= \sqrt2 \int_0^\infty \left[\int_0^{x^2} \frac{dy}{1+y^2}\right] \, \frac{dx}{x^2} \\ &= \sqrt2 \int_0^\infty \left[\int_{\sqrt y}^\infty \frac{dx}{x^2}\right] \, \frac{dy}{1+y^2} \\ &= \sqrt2 \int_0^\infty \frac{dy}{\sqrt y\left(1+y^2\right)} \\ &= 2\sqrt2 \int_0^\infty \frac{dy}{1+y^4} \tag2\\ &= 2\sqrt2\cdot\frac\pi{2\sqrt2} = \boxed\pi \end{align*}$$

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• $(1)$ : substitute $x\mapsto\dfrac1{\sqrt2\,x}$
• $(2)$ : $y\mapsto y^2$