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Find the value of the following definite integral:

$$\int_0^{1}\frac{\arctan(t)}{1+t}dt$$

I tried using integration by parts, but it gives another complicated integral, so possibly the antiderivative does not exist. What other methods can I apply?

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    $\begingroup$ Are you sure that the denominator is not $1+t^2$ instead of $1+t$ ? $\endgroup$ Oct 7, 2014 at 9:40
  • $\begingroup$ An antiderivative exists according to wolframalpha, but it's weird. Better try with definite integration methods. Where did you find it? Are you sure it has a closed form? $\endgroup$
    – UserX
    Oct 7, 2014 at 9:41
  • $\begingroup$ @Claude, yes of course. Otherwise the question is just a simple substitution $\endgroup$
    – Shubham
    Oct 7, 2014 at 9:43
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    $\begingroup$ $\arctan\dfrac\pi4$ doesn't make much sense. Perhaps you meant to write $\displaystyle\int_0^1$, since $\arctan1=\dfrac\pi4$ ? In which case, the answer is $\dfrac\pi8\ln2$. $\endgroup$
    – Lucian
    Oct 7, 2014 at 9:56
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    $\begingroup$ This integral is so difficult that I just wondered. $\endgroup$ Oct 7, 2014 at 9:58

2 Answers 2

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Let us write $$I=\int_0^1\dfrac{\arctan t}{1+t}dt$$ The change of variables $t=\dfrac{1-x}{1+x}$ shows that $$\eqalign{I&=\int_0^1\arctan\left(\frac{1-x}{1+x}\right)\frac{dx}{1+x}\\ &=\int_0^1\left(\frac{\pi}{4}-\arctan x\right)\frac{dx}{1+x}\\ &=\frac{\pi}{4}\int_0^1\frac{dx}{1+x}-\int_0^1\frac{\arctan x}{1+x}dx\\ &=\frac{\pi}{4}\ln 2-I }$$ Thus $I=\dfrac{\pi}{8}\ln 2$.$\qquad \square$

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  • $\begingroup$ Don't the limits on the integral have to change when switching to x? $\endgroup$
    – Ryan
    Oct 7, 2014 at 14:35
  • $\begingroup$ @Ryan They did, $0$ becomes $1$ and $1$ becomes $0$ and you have a minus sign when you go from $dt$ to $dx$... $\endgroup$ Oct 7, 2014 at 15:00
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It is not as easy an integral as it first appears to be. I am cutting/pasting solutions in four cases obtained by Mathematica Version 8. ...including the upper limit 1 you changed.

First case result is real, even if it involves $i$ but fails to annul $i$ in the final expression.

Second case is what you asked for, evaluated numerically.

Third one actually involves Arctan, Log, and Catalan constant.

Fourth case with upper limit 1 has an easily obtainable result, but connection to simplification due the upper limit 1 is surprising...

$ \text{Integrate}[\text{ArcTan}[(u)]/(1+u),u] $

$ 1/32 (-5 I[\text{Pi}]{}^{\wedge}2+8 I[\text{Pi}] \text{ArcTan}[u]-32 I \text{ArcTan}$

$[u]{}^{\wedge}2-24[\text{Pi}] \text{Log}[2]+16[\text{Pi}] \text{Log}$

$[1+E{}^{\wedge}(-2 I \text{ArcTan}[u])] $

$-32 \text{ArcTan}[u] \text{Log} [1+E{}^{\wedge}(-2 I \text{ArcTan}[u])]+ $

$ 8[\text{Pi}] \text{Log}[1-I E{}^{\wedge}(2 I \text{ArcTan}[u])] $

$ +32 \text{ArcTan}[u] \text{Log}[1-I E{}^{\wedge}(2 I \text{ArcTan}[u])]+ $

$8[\text{Pi}] \text{Log}[1+u{}^{\wedge}2]-8[\text{Pi}] \text{Log}$ $[\text{Sin}$

$[[\text{Pi}]/4+ \text{ArcTan}[u]]]-16 I \text{PolyLog}[2,-E{}^{\wedge}(-2 I $

$\text{ArcTan}[u])]-16 I \text{PolyLog}[2,I E{}^{\wedge}(2 I \text{ArcTan}[u])]) $

$ \text{NIntegrate}[\text{ArcTan}[(u)]/(1+u),\{u,0,\text{Pi}/4.\}] $

0.189728

$ \text{Integrate}[\text{ArcTan}[(u)]/(1+u),\{u,0,\text{Pi}/4\}] $

$ 1/96 (-48 \text{Catalan}-47 I[\text{Pi}]{}^{\wedge}2+[\text{Pi}] (24 I$

$ \text{ArcTan}[[\text{Pi}]/4]-2$

$\text{Log}[4096]+24 (\text{Log}[1-I E{}^{\wedge}(2 I \text{ArcTan}$

$[[\text{Pi}]/4])]+2 \text{Log}[I/(-4 I+[\text{Pi}])]+\text{Log}[16+[\text{Pi}]$

${}^{\wedge}2]+\text{Log}[ \text{Csc}[[\text{Pi}]/4+ $

$ \text{ArcTan} [[\text{Pi}]/4]]]))-48 I (2 \text{ArcTan}[[\text{Pi}]/4] $

$ {}^{\wedge}2+2I \text{ArcTan}[[\text{Pi}]/4] (\text{Log}[-((1+I)/(4 $

$ I+[\text{Pi}]))]-\text{Log}[(8 I)/((-4 I+[\text{Pi}]) $

$ (4+[\text{Pi}]))])+\text{PolyLog}[2,-((4+I[\text{Pi}])/(4 $

$ I+[\text{Pi}]))]+\text{PolyLog}[2,(4 I+[\text{Pi}])/(-4 I+[\text{Pi}])])) $

$ \text{Integrate}[\text{ArcTan}[(u)]/(1+u),\{u,0,1\}] $

$ log (2) \pi/8 $

The given function is regular, is a product of two functions one increasing, other decreasing, so has a maximum, ( at t = 1.22913).

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  • $\begingroup$ In Mathematica, you can choose to copy output as Latex code. This makes it much more human-readable. $\endgroup$
    – user111187
    Oct 7, 2014 at 15:14
  • $\begingroup$ Mathematica forever? Not quite in this case :p $\endgroup$
    – yo'
    Oct 7, 2014 at 16:10
  • $\begingroup$ Sorry, messed up with it first time me doing.. At least serves the purpose that not so nice integration limits invite Arctan Log, PolyLog, Catalan and what have you :) $\endgroup$
    – Narasimham
    Oct 7, 2014 at 16:15

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