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I am wondering if there is a way to integrate the function $$\int_{0}^∞ \frac{\tan^{-1}(\pi{x})-\tan^{-1}(x)}{e^{x^2}}$$ without resorting to Simpson's rule/numerical integration. My first thought was to split it into two integrals and use integration by parts on each, but that did not work. Each step integrating by parts just convoluted it more and more.

Depending how you integrate by parts, you either get the chain of convoluted integrals I did, or immediately wind up with the Gauss Error Function $\text{erf}(x)$.

I'm self-taught and still relatively new, so please forgive me if I missed something simple. Anyone know what to do here?

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  • $\begingroup$ Can you write out, as part of your question, your $erf(x)$ result? $\endgroup$ – The Count Apr 18 at 3:50
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This is a real monster ... but a CAS gave a result I am sure you will enjoy !

If $$I_a=\int_0^\infty e^{-x^2} \tan ^{-1}(a x)\,dx$$ then, assuming that $a$ is a real positive number, $$I_a=\frac{1}{4} \sqrt{\pi } \left(\text{erfi}\left(\frac{1}{a}\right) (\gamma -2 \log (a))+\pi \right)+$$ $$\frac{\text{HypergeometricPFQ}^{(\{0,0\},\{0,1\},0)}\left(\left\{\frac{1}{2},1\right\},\left\{\frac{3}{2},1\right\},\frac{1}{a^2}\right)-2 \, _2F_2\left(\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2};\frac{1}{a^2}\right)} {2 a} $$where appear hypergeometric functions and derivatives.

Its decimal representation of $(I_\pi-I_1)$ is $0.372711790424249$

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  • $\begingroup$ Thank you Claude! Is there any way you could explain in detail how CAS got there? I really want to know what steps are involved/how many. $\endgroup$ – Rob Churan Apr 18 at 20:29
  • $\begingroup$ @RobChuran. Me too, be sure ! $\endgroup$ – Claude Leibovici Apr 19 at 8:49
  • $\begingroup$ Hello again Claude. Just wondering what CAS you used? I've gone through a few that go kaput with my "monster." It'd be nice to play with the same one myself. $\endgroup$ – Rob Churan Apr 24 at 6:09
  • $\begingroup$ @RobChuran. For this one, I used the Wolfram Development Platform. It is free; you just need to register. $\endgroup$ – Claude Leibovici Apr 24 at 6:53
  • $\begingroup$ Thank you, I'll check it out! $\endgroup$ – Rob Churan Apr 25 at 1:19

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