7
$\begingroup$

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?

$\endgroup$
8
  • 1
    $\begingroup$ Are you sure that the denominator is not $1+t^2$ instead of $1+t$ ? $\endgroup$
    – Claude Leibovici
    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
  • 1
    $\begingroup$ $\arctan\dfrac\pi4$ 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
  • 1
    $\begingroup$ This integral is so difficult that I just wondered. $\endgroup$
    – Claude Leibovici
    Oct 7, 2014 at 9:58

2 Answers 2

Reset to default
12
$\begingroup$

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 $I=\dfrac{\pi}{8}\ln 2$.$\qquad \square$

$\endgroup$
2
  • $\begingroup$ 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 $dx$... $\endgroup$
    – Omran Kouba
    Oct 7, 2014 at 15:00
1
$\begingroup$

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).

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.