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I'm trying to solve $$\int_{0}^{\infty} \frac{1}{x^2}\int_{0}^{x}\frac{\arctan\left(\frac{x}{\sqrt{t^2+2}}\right)}{(t^2+1)\sqrt{t^2+2}}\, dt \, dx $$

The value should be $\displaystyle \frac{\pi}{2}\ln(2)$. I tried different substitutions, but it doesn't work out. Changing the integration order seems not really nice as below: $$\int_1^{\infty} \frac{2\sqrt{z+1}\arctan\left(\sqrt{\frac{z-1}{z+1}}\right)+\sqrt{z-1}(\ln(2z)-\ln(z-1))}{4z(z-1)(z+1)} dz$$

Has anyone a good start for me? Maybe Feynman? Thanks!

Lee David Chung Lin
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DavidP
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    Can you mention the source of the problem? Like how do you know what the answer should be? – Zacky Feb 13 '20 at 17:03
  • I was reading it somewhere at internet. I'm not sure the solution is right, because all my attempts calculating it failed. The source of the problem is just fun - I like integrals. :) – DavidP Feb 13 '20 at 17:12
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    Where is somewhere? It's nice if you like integrals, but others do too. So mentioning the source would enable others to find perhaps more integrals :) As a start for this integral, change the order of the integration. – Zacky Feb 13 '20 at 17:16
  • Thanks! I will try. The source is a german math forum - it was asked 2012, but with no solution yet. You want the link? – DavidP Feb 13 '20 at 17:51
  • @Zacky If my calculation is right changing the order leads to $\int_1^{\infty} \frac{2\sqrt{z+1}\arctan\left(\sqrt{\frac{z-1}{z+1}}\right)+\sqrt{z-1}(\ln(2z)-\ln(z-1))}{4z(z-1)(z+1)} dz$ Seems not really nice - but I'll think about it tomorrow. Thanks! – DavidP Feb 13 '20 at 19:39
  • Thank you, if the forum is not in English, then not really. I have no idea how you approached it after changing the order of the integrals, so I can't say anything unless you include your work to arrive there in the question body. – Zacky Feb 13 '20 at 19:59
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    I might write an answer if I can reduce some of the steps, but I obtained successively: $$I=\frac12\int_0^\infty \frac{1}{(1+t^2)(2+t^2)}\int_0^1 \ln\left(1+\frac{1+\frac{2}{t^2}}{z^2}\right)dzdt$$ $$=\frac{\pi}{2}\int_0^1 \left(\ln\left(\sqrt{1+\frac{1}{z^2}}+\frac{\sqrt 2}{z}\right)-\frac{1}{\sqrt 2}\ln\left(\sqrt{1+\frac{1}{z^2}}+\frac{1}{z}\right)\right)dz$$ The last inner integral can be computed through IBP in order to obtain $\ln 2$ and the result would be confirmed. But looking at it, seems like I reversed the order of integration twice.. so perhaps it wasn't necessary at all. – Zacky Feb 13 '20 at 20:00

1 Answers1

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Suppose $\displaystyle\int\limits_0^x\frac{\arctan\left(\frac{x}{\sqrt{t^2+2}}\right)}{(t^2+1)\sqrt{t^2+2}}\, dt$. It's well known that $\arctan z=\dfrac{\pi}{2}-\text{arccotan} z$, so our integral is $$\dfrac{\pi}{2}\int\limits_0^x\frac{dt}{(t^2+1)\sqrt{t^2+2}}\, dt-\int\limits_0^x\frac{\text{arccotan}\left(\frac{x}{\sqrt{t^2+2}}\right)}{(t^2+1)\sqrt{t^2+2}}\, dt.$$

It's obvious that $\dfrac{1}{z}\text{arccotan}\dfrac{x}{z}=\dfrac{\pi}{2z}-\displaystyle\int_0^x\dfrac{dy}{y^2+z^2}$. Given that $z=\sqrt{2+t^2}$ and substituting the last equality in second integral, we get that our two integrals turns into $$\displaystyle\int\limits_0^x\int\limits_0^x\dfrac{dydt}{(t^2+1)(y^2+t^2+2)}=I.$$ Further, it's easy to see that $I=\displaystyle\int\limits_0^x\int\limits_0^x\dfrac{dydt}{(t^2+1)(y^2+1)}-\displaystyle\int\limits_0^x\int\limits_0^x\dfrac{dydt}{(y^2+1)(y^2+t^2+2)}$.

Since in the second integral everything is symmetric, $\displaystyle\int\limits_0^x\int\limits_0^x\dfrac{dydt}{(y^2+1)(y^2+t^2+2)}=I$ (we can also make a replacement $t\to y$ and $y\to t$). Thus $$I=\dfrac{1}{2}\displaystyle\int\limits_0^x\int\limits_0^x\dfrac{dydt}{(t^2+1)(y^2+1)}=\dfrac{1}{2}\left(\displaystyle\int\limits_0^x\dfrac{dt}{t^2+1}\right)^2=\dfrac{\arctan^2x}{2}.$$

It remains to calculate the integral $\dfrac{1}{2}\displaystyle\int\limits_0^\infty\dfrac{\arctan^2x}{x^2}dx=\displaystyle\int\limits_0^\infty\dfrac{\arctan x}{x(1+x^2)}dx$ (here we applying integrate by parts). Can you finish it yourself? Hint: you can consider the integral $\displaystyle\int\limits_0^\infty\dfrac{\arctan ax}{x(1+x^2)}dx$ and differentiate it by the parameter.

amWhy
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thing
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  • Thanks a lot - nice! The last integral can also be solved by complex Analysis. See there: https://math.stackexchange.com/questions/2401761/compute-int-0-pi-2-frac-x-tan-x-dx-using-contour-integration – DavidP Feb 16 '20 at 11:11