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How to compute the integral

$$\int_0^1 {\tan^{-1}\left(x\right) \over {1+x}}\,{\rm d}x$$

Felix Marin
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DebojyotiMukh
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2 Answers2

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Use the substitution $x=\frac{1-y}{1+y}$ so that $\mathrm{d}x=-\frac{2\mathrm{d}y}{(1+y)^2}$ and $\tan^{-1}(x)+\tan^{-1}(y)=\frac\pi4$. $$ \begin{align} \int_0^1\frac{\tan^{-1}(x)}{1+x}\,\mathrm{d}x &=\int_0^1\frac{\frac\pi4-\tan^{-1}(y)}{\frac2{1+y}}\frac{2\mathrm{d}y}{(1+y)^2}\\ &=\int_0^1\frac{\frac\pi4-\tan^{-1}(y)}{1+y}\,\mathrm{d}y\\[4pt] &=\frac\pi4\log(2)-\int_0^1\frac{\tan^{-1}(y)}{1+y}\,\mathrm{d}y\\ \end{align} $$ Therefore, $$ \int_0^1\frac{\tan^{-1}(x)}{1+x}\,\mathrm{d}x=\frac\pi8\log(2) $$

robjohn
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    Shouldn't the first integral in y be: $\int_1^0\frac{\frac\pi4-\tan^{-1}(y)}{\frac2{1+y}}\frac{2\mathrm{d}y}{(1+y)^2}$ ?? – K. Rmth Nov 09 '13 at 22:52
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    @K.Rmth: except for the minus sign (in the $\mathrm{d}x$) and the reversal of the limits. – robjohn Nov 09 '13 at 22:53
  • +1. I see the substitution $x=\frac{1-y}{1+y}$ made frequently with nasty integrals, sort of like the Weierstrass substitution. What sort of things do you look for in an integral to think about using your substitution? – Joe Nov 09 '13 at 23:16
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    @Joe: In this case, I started with $\tan^{-1}(x)+\tan^{-1}(y)=\frac\pi4$ and went from there. I can't recall which answer it was, but I've used this substitution before. – robjohn Nov 09 '13 at 23:24
  • @robjohn Okay, thanks. I'm interested in learning more of the non-canonical tricks when it comes to integrals that come up from Putnam, USAMO, and such and I've seen the substitution a few times before. – Joe Nov 09 '13 at 23:53
  • Nice. UPVote $0$k. – Felix Marin Nov 10 '13 at 00:09
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    @Joe: another nice thing about this substitution, that is not used here, is that not only does $\frac{\mathrm{d}x}{1+x}=-\frac{\mathrm{d}y}{1+y}$, but also $\frac{\mathrm{d}x}{1+x^2}=-\frac{\mathrm{d}y}{1+y^2}$ – robjohn Nov 10 '13 at 01:28
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$\int_0^1 {{tan^{-1}x}\over {1+x}}dx=[tan^{-1}x\int {dx\over{1+x}}]_0^1-\int_0^1{log(1+x)\over{1+x^2}}dx={\pi\over 8}log2$

Putting $x=tan y$ in the 2nd integral

DebojyotiMukh
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