How to evaluate $$I=\int_{0}^{\infty}\arctan (x^2)\sin(x^2)\mathrm dx$$ with the help of Wolfram alpha,I got the answer below $$I=\frac{\pi^{2/3}\text{erfc(1)}(\text{erfi(1)}+1)}{4\sqrt2}$$ But I don't know how to prove it.

  • $\begingroup$ I am ready to bet that this involves the Dawson integral. But, how to prove it, this is the question. $\endgroup$ – Claude Leibovici Jun 6 '16 at 13:04

A simple approach is to use differentiation under the integral sign. If we set: $$ I(\alpha)=\int_{0}^{+\infty}\arctan(\alpha x^2)\sin(x^2)\,dx $$ for any $\alpha> 0$, we have: $$ I'(\alpha) = \int_{0}^{+\infty}\frac{x^2}{1+a^2 x^4}\,\sin(x^2)\,dx $$ that can be computed trough the Laplace transform: $$ I'(\alpha) = \frac{\pi e^{-1/\alpha}}{2\sqrt{2\pi^3}}\left(\text{Erfi}\left(\frac{1}{\sqrt{\alpha}}\right)-e^{2/\alpha}\text{Erfc}\left(\frac{1}{\sqrt{\alpha}}\right)\right) $$ and by: $$ I(1)=\int_{0}^{1}I'(\alpha)\,d\alpha $$ the claim follows.


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