8
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
8
+50
$\begingroup$

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. $\qquad \qquad \Box$

$\endgroup$
  • $\begingroup$ 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
7
$\begingroup$

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

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

or

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

$\endgroup$
  • $\begingroup$ 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

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.