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How can I prove the following identity? $$\int_0^1\frac{x^2-2\,x+2\ln(1+x)}{x^3\,\sqrt{1-x^2}}\mathrm dx=\frac{\pi^2}8-\frac12$$

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  • $\begingroup$ Just curious (for my own learning's sake), what sort of class involves integrals like this? It's certainly above anything in my Calc II course I took. Is this like Real Analysis (or perhaps Complex Analysis)? $\endgroup$ – apnorton Nov 7 '13 at 23:55
  • $\begingroup$ How did you get the answer? $\endgroup$ – Mhenni Benghorbal Nov 8 '13 at 0:13
  • $\begingroup$ Where did this came from? What's the history behind this identity? $\endgroup$ – Lucas Zanella Nov 8 '13 at 2:53
  • $\begingroup$ @MhenniBenghorbal It was part of the problem. Anyways, it would be easy to guess using a numeric approximation and wolframalpha.com $\endgroup$ – Laila Podlesny Nov 8 '13 at 23:09
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First observe that

$$x^2-2 x+2 \log{(1+x)} = 2 \sum_{k=3}^{\infty} (-1)^{k+1} \frac{x^k}{k}$$

The integral is then equal to

$$2 \sum_{k=0}^{\infty} \frac{(-1)^k}{k+3} \int_0^1 dx \frac{x^k}{\sqrt{1-x^2}}$$

Now, we will need separate treatments for the even and odd terms (1):

$$\int_0^1 dx \frac{x^k}{\sqrt{1-x^2}} = \begin{cases} \frac{\displaystyle 1}{\displaystyle 2^{2 k}} \displaystyle \binom{2 k}{k} \frac{\pi}{2} & k \: \text{even}\\ \frac{\displaystyle 2^{2 k-1}}{\displaystyle k \binom{2 k}{k}} & k \: \text{odd} \end{cases} $$

That is, the integral is now equal to the difference between two sums:

$$ \pi \sum_{k=0}^{\infty} \frac{1}{2 k+3} \frac{1}{2^{2 k}} \binom{2 k}{k} - \frac12 \sum_{k=1}^{\infty} \frac{1}{ k+1} \frac{\displaystyle 2^{2 k}}{\displaystyle k \binom{2 k}{k}}$$

We now evaluate each sum in turn. For the first, let

$$f(x) = \sum_{k=0}^{\infty} \frac{1}{2 k+3} \frac{1}{2^{2 k}} \binom{2 k}{k} x^{2 k+3} $$

Then

$$f'(x) = x^2 \sum_{k=0}^{\infty} \frac{1}{2^{2 k}} \binom{2 k}{k} x^{2 k} = \frac{x^2}{\sqrt{1-x^2}}$$

which means that, enforcing the condition that $f(0)=0$ (2),

$$f(x) = \int dx \frac{x^2}{\sqrt{1-x^2}} = \frac{1}{2} \arcsin(x)-\frac{1}{2} x \sqrt{1-x^2}$$

The sum in question is equal to $f(1) = \pi/4$. For the second sum, define

$$g(x) = \sum_{k=1}^{\infty} \frac{1}{k( k+1)} \frac{\displaystyle 2^{2 k}}{\displaystyle \binom{2 k}{k}} x^{k+1}$$

Then (see this answer for a reference)

$$g''(x) = \frac{1}{x} \sum_{k=1}^{\infty} \frac{(4 x)^k}{\displaystyle \binom{2 k}{k}} = \frac{\displaystyle 1+\frac{ \arcsin\left(\sqrt{x}\right)}{\sqrt{x(1-x)}}}{1-x}$$

Integrating twice and enforcing the condition that $g(0)=0$ and $g'(0)=0$, we find that (3)

$$g(x) = x+\arcsin\left(\sqrt{x}\right)^2-2 \sqrt{x(1-x)} \arcsin\left(\sqrt{x}\right) $$

The second sum is then

$$g(1) = 1+\frac{\pi^2}{4}$$

The value of the integral we seek is then equal to

$$\pi f(1) - \frac12 g(1) = \pi \frac{\pi}{4} - \frac12 \left ( 1+ \frac{\pi^2}{4} \right ) = \frac{\pi^2}{8} - \frac12$$

as was to be shown.

ADDENDUM

I think I should fill in some gaps of the above proof. I will go through each intermediate result in turn so that the solution is more self-contained. The integrals I evaluate here are not as difficult as they appear, although there is one subtlety that should be pointed out.

Equation (1)

$$\int_0^1 dx \frac{x^k}{\sqrt{1-x^2}}$$

a) $k$ even, i.e., $k=2 m$, $m \in \{0,1,2,\ldots\}$

Sub $x=\sin{t}$ to see that this integral is equal to

$$I_m = \int_0^{\pi/2} dt \, \sin^{2 m}{t} $$

Integrate by parts to see that

$$\begin{align}I_m &= -\underbrace{\left [ \cos{t} \sin^{2 m-1}{t} \right ]_0^{\pi/2}}_{\text{this}=0} + (2 m-1) \underbrace{\int_0^{\pi/2} dt \, \cos^2{t} \sin^{2 m-2}{t}}_{\cos^2{t}=1-\sin^2{t}}\\ &= (2 m-1) I_{m-1} - (2 m-1) I_m\end{align}$$

Thus,

$$I_m = \frac{2 m-1}{2 m} I_{m-1} = \frac{(2 m-1)(2 m-3)\cdots (3)(1)}{(2 m)(2 m-2)\cdots (2)} I_0$$

where $I_0 = \int_0^{\pi/2} dt = \pi/2$. We may rearrange the above result by multiplying the numerator by the denominator, and we have for even values of $k$:

$$I_m = \frac{1}{2^{2 m}} \binom{2 m}{m} \frac{\pi}{2} $$

b) $k$ odd, i.e., $k=2 m+1$, $m \in \{0,1,2,\ldots\}$

We perform identical manipulations as above, but now we get that

$$I_m = \frac{(2 m)(2 m-2)\cdots (2)}{(2 m+1)(2 m-1)\cdots (3)} I_1 $$

where $I_1 = \int_0^{\pi/2} dt \, \sin{t} = 1$. Using similar manipulations as above (except we multiply the denominator by the numerator), we have

$$I_m = \frac{1}{2 m+1} \frac{2^{2 m}}{\displaystyle \binom{2 m}{m}} $$

You may note, however, that this is not the result I displayed in the proof. Good reason: this form would complicate the series approach to evaluating the sum. To this effect, let's map $m \mapsto m-1$ and consider $m \in \{1,2,3,\ldots\}$. Then

$$I_m = \frac{2^{2 m-2}}{2 m-1} \frac{[(m-1)!]^2}{(2 m-2)!} = \frac{2^{2 m-1}}{\displaystyle m \binom{2 m}{m}} $$

as asserted.

Equation (2)

$$\underbrace{\int dx \frac{x^2}{\sqrt{1-x^2}}}_{x=\sin{t}} = \int dt \, \sin^2{t} = \frac{t}{2} - \frac12 \sin{t} \cos{t}$$

form which the posted result follows.

Equation (3)

Here we have 2 integrations. First,

$$g'(x) = \underbrace{\int dx \frac{1+\frac{\arcsin{\sqrt{x}}}{\sqrt{x (1-x)}}}{1-x}}_{x=u^2} = \underbrace{2 \int du \, \frac{u + \frac{\arcsin{u}}{\sqrt{1-u^2}}}{1-u^2}}_{u=\sin{t}} = 2 \int dt \, \tan{t} + 2 \int dt \, t \sec^2{t} $$

Do the second integral by parts:

$$2 \int dt \, t \sec^2{t} = 2 t \tan{t} - 2 \int dt \, \tan{t}$$

Thus we have a fortuitous cancellation, and using $t=\arcsin{\sqrt{x}}$, and enforcing $g'(0)=0$, we have

$$g'(x) = 2 \sqrt{\frac{x}{1-x}}\arcsin{\sqrt{x}}$$

So, second, we must integrate this result to get $g(x)$. We use similar substitutions as above (i.e., $x=u^2$, $u=\sin{t}$):

$$g(x) = 4 \int du \, \frac{u^2}{\sqrt{1-u^2}} \arcsin{u} = 4 \int dt \, t \, \sin^2{t}$$

Now, integrate by parts:

$$4 \int dt \, t \, \sin^2{t} = 2 t (t - \sin{t} \cos{t}) - 2 \int dt \, (t - \sin{t} \cos{t}) = t^2 - 2 t \sin{t} \cos{t} + \sin^2{t} +C $$

Now, use $t = \arcsin{\sqrt{x}}$ and the fact that $g(0)=0$ and get

$$g(x) = \arcsin{\left ( \sqrt{x}\right )}^2 - 2 \sqrt{x (1-x)} \arcsin{\left ( \sqrt{x}\right )} + x$$

as posted above.

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    $\begingroup$ Nice solution. Ron Gordon $\endgroup$ – juantheron Nov 8 '13 at 2:55
  • $\begingroup$ Bravo again Ron! +1 $\endgroup$ – Bennett Gardiner Nov 8 '13 at 12:55
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    $\begingroup$ @BennettGardiner: thanks, as always, you show me so much kindness. $\endgroup$ – Ron Gordon Nov 8 '13 at 14:29
  • $\begingroup$ Excellent work! :) $\endgroup$ – Ahaan S. Rungta Nov 9 '13 at 1:22
  • $\begingroup$ Out of curiosity, why is the integral then equal to the difference between the two sums? Because of the subtraction of those terms we "picked out" by treating it by cases of $k$ being even and odd? $\endgroup$ – jm324354 Nov 18 '14 at 23:11

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