# Prove $\int_0^1\frac{x^2-2\,x+2\ln(1+x)}{x^3\,\sqrt{1-x^2}}dx=\frac{\pi^2}8-\frac12$

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

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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)? –  anorton Nov 7 '13 at 23:55
How did you get the answer? –  Mhenni Benghorbal Nov 8 '13 at 0:13
Where did this came from? What's the history behind this identity? –  Lucas Zanella Nov 8 '13 at 2:53
@MhenniBenghorbal It was part of the problem. Anyways, it would be easy to guess using a numeric approximation and wolframalpha.com –  Laila Podlesny Nov 8 '13 at 23:09

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.

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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Nice solution. Ron Gordon –  juantheron Nov 8 '13 at 2:55
Bravo again Ron! +1 –  Bennett Gardiner Nov 8 '13 at 12:55
@juantheron: thanks. I've been meaning to ask: is juantheron a play on Juan Perón, or is it just your name and I'm being too imaginative? –  Ron Gordon Nov 8 '13 at 14:29
@BennettGardiner: thanks, as always, you show me so much kindness. –  Ron Gordon Nov 8 '13 at 14:29
Excellent work! :) –  Ahaan Rungta Nov 9 '13 at 1:22