Something is wrong with this integral (in terms of splitting them out)

$$\int_{0}^{1}{2x^2-2x+\ln[(1-x)(1+x)^3]\over x^3\sqrt{1-x^2}}\mathrm dx=\color{blue}{-1}\tag1$$

My try:

Splitting the integral

$$\int_{0}^{1}{2x^2-2x\over x^3\sqrt{1-x^2}}\mathrm dx+\int_{0}^{1}{\ln[(1-x)(1+x)^3]\over x^3\sqrt{1-x^2}}\mathrm dx=I_1+I_2\tag2$$

Note that $I_1$ and $I_2$ diverge, so how can we tackle it as a whole?

  • $\begingroup$ maybe using the change of variable $y=\sqrt{\dfrac{1-x}{1+x}}$ and then, using the change of variable $z=\dfrac{1-y}{1+y}$ $\endgroup$ – FDP May 8 '17 at 8:09
  • $\begingroup$ I try these change of variables, it looks ugly, believe me! I am sure (because of such a simple closed form) there must be an easy way of simplifying this integral. $\endgroup$ – gymbvghjkgkjkhgfkl May 8 '17 at 10:59
  • $\begingroup$ It's not ugly at all. Only x^3 left for the denominator according to Maxima. Anyway, you have, as usual, build this integral from another one, you know the method you have used. $\endgroup$ – FDP May 8 '17 at 11:21
  • $\begingroup$ After the two changes of variable, $-\tfrac{1}{2}\int_{0}^{1}\frac{-\mathrm{log}\left( 1-x\right) +2\cdot x+\left( -2\cdot \mathrm{log}\left( 1-x\right) -4\right) \cdot {{x}^{2}}+2\cdot {{x}^{3}}-\mathrm{log}\left( 1-x\right) \cdot {{x}^{4}}+\left( -3\cdot {{x}^{4}}-6\cdot {{x}^{2}}-3\right) \cdot \mathrm{log}\left( x+1\right) +\left( 2\cdot {{x}^{4}}+4\cdot {{x}^{2}}+2\right) \cdot \mathrm{log}\left( {{x}^{2}}+1\right) }{{{x}^{3}}}dx$ $\endgroup$ – FDP May 8 '17 at 11:23
  • $\begingroup$ No time, for now, to terminate this computation. $\endgroup$ – FDP May 8 '17 at 11:25

I will present an answer without Gamma functions and without series' expansions. Instead, a more general problem is solved, where the OP's question is a special case.

Let $$ I(a) = \int_{0}^{1}{2a^2x^2-2ax+\ln[(1-ax)(1+ax)^3]\over x^3\sqrt{1-x^2}}\mathrm dx $$

The answer to the OP's question is then given by $I(a = 1)$.

Then, after partial differentiation w.r.t. $a$,

$$ I'(a) = \int_{0}^{1} \frac{2a^2(2ax - 1)}{(a^2 x^2 - 1)\sqrt{1-x^2}}\mathrm dx $$

Note that this removes the problematic factor $x^3$ in the denominator - this issue was solved in the previous solution by series' expansion of the $\log$. Hence, convergence is established and the order of integrations $x,a$ can be exchanged.

The $x$-integration gives $$ I'(a) = a^2 \Big[ \frac{2 \arctan(\frac{x \sqrt{1 - a^2}}{\sqrt{1 - x^2}})}{\sqrt{1 - a^2}} + \frac{4\arctan(\frac{a \sqrt{1 - x^2}}{\sqrt{1 - a^2}})}{\sqrt{1 - a^2}} \Big]_{x=0}^{1} $$

which is

$$ I'(a) = a^2 \Big[ \frac{\pi}{\sqrt{1 - a^2}} - \frac{4\arctan(\frac{a }{\sqrt{1 - a^2}})}{\sqrt{1 - a^2}} \Big] = a^2 \Big[ \frac{\pi}{\sqrt{1 - a^2}} - \frac{4\arcsin({a })}{\sqrt{1 - a^2}} \Big] $$

Now we integrate w.r.t. $a$: $$ I(a) = \int I'(a) \rm{d} a = \int a^2 \Big[ \frac{\pi}{\sqrt{1 - a^2}} - \frac{4\arcsin({a })}{\sqrt{1 - a^2}} \Big] \rm{d} a $$

Replacing $a = \sin (y)$ gives

$$ I(y) = \int \sin^2(y) \Big[ {\pi} - {4 y}\Big] \rm{d} y = y \sin(2y) + (\pi y)/2 - \sin^2(y) - y^2 - (\pi \sin(2y))/4 + C $$

We need the constant $C$. Obviously $0 = I(a=0) = I(y=0)$ which gives $C=0$.

The function in question by the OP is $I(a=1) = I(y=\pi/2) = -1$.

This solves the OP'S question. $\qquad \qquad \Box$

  • $\begingroup$ Nice answer thanks ! $\endgroup$ – FDP May 9 '17 at 17:41
  • $\begingroup$ good one .... +1 $\endgroup$ – tired May 11 '17 at 6:40
  • $\begingroup$ Wow! An extra bounty on that one - thanks! $\endgroup$ – Andreas May 14 '17 at 9:55

You have to split as

$$ I=\int_0^1\frac{x^2/2+x+\log(1-x)}{x^3\sqrt{1-x^2}}+\int_0^1\frac{3x^2/2-3x+3\log(1+x)}{x^3\sqrt{1-x^2}}=\color{red}{I_1}+\color{blue}{I_2} $$

Now we can calculate

$$ \color{red}{I_1}=-\sum_{n=3}^{\infty}\frac{1}{n}\int_0^1\frac{x^{n-3}}{\sqrt{1-x^2}}=-\frac{\sqrt{\pi}}{2}\sum_{n=3}^{\infty}\frac{1}{n}\frac{\Gamma\left(\frac n2-1\right)}{\Gamma\left(\frac n2-\frac12\right)}\underbrace{=}_{(1)}\\ -\frac{1}{16}\sum_{n=3}^{\infty}\frac{2^n}{n}\frac{\Gamma^2\left(\frac n2-1\right)}{\Gamma\left(n-1\right)}=-\frac12\int_0^1dtt\sum_{k=1}^{\infty}(2t)^k\frac{ \Gamma^2 \left(\frac k2\right)}{k!}\underbrace{=}_{(2)}\\ \int_0^1dtt\arcsin(t)(\pi+\arcsin(t))=\color{red}{-\frac{1}{4}+\frac{3\pi^2}{16}} $$

where we have used Legendre's duplication formula in $(1)$ and the results from this nice question in $(2)$ (the last integration is trivial after employing $t=\sin(q)$ so i spare it here).

Playing exactly the same game with $I_2$ we obtain

$$ \color{blue}{I_2}=-3\int_0^1dtt\arcsin(t)(\pi-\arcsin(t))=\color{blue}{-\frac{3}{4}-\frac{3\pi^2}{16}} $$


$$ I=\color{red}{I_1}+\color{blue}{I_2}=-1 $$

  • $\begingroup$ Great answer (+1), I thought the proof would be very lengthy. $\endgroup$ – gymbvghjkgkjkhgfkl May 8 '17 at 20:16
  • $\begingroup$ I got a solution without series' expansions and without Gamma functions. See below. $\endgroup$ – Andreas May 9 '17 at 13:12
  • $\begingroup$ What is $dtt$ ? $\endgroup$ – Zaid Alyafeai May 13 '17 at 1:37
  • 1
    $\begingroup$ @ZaidAlyafeai the differential $dt$ multiplied by $t$ $\endgroup$ – tired May 14 '17 at 16:19

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