TITLE: Prove $\int_0^1\frac{x^2-2\,x+2\ln(1+x)}{x^3\,\sqrt{1-x^2}}\mathrm dx=\frac{\pi^2}8-\frac12$
QUESTION [35 upvotes]: How can I prove the following identity?
$$\int_0^1\frac{x^2-2\,x+2\ln(1+x)}{x^3\,\sqrt{1-x^2}}\mathrm dx=\frac{\pi^2}8-\frac12$$

REPLY [51 votes]: 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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