Integrate the following: $\int \frac {\log x}{\sqrt {1-x^2}}\,dx$

Question

What I tried :

$$\int \frac {\log x}{\sqrt {1-x^2}} dx$$

$$= \log x \int \frac {1}{\sqrt {1-x^2}} dx - \int \frac{1}{x} \left(\int \frac {1}{\sqrt {1-x^2}} \, dx\right) \, dx$$

Now,

$$\int \frac {1}{\sqrt {1-x^2}} \, dx = \arcsin x + c$$

But inputting this into the $\displaystyle\int \frac{1}{x} \int \left(\frac {1}{\sqrt {1-x^2}} \, dx\right) \,dx$ part is not getting me into a neat solution. Looks like there is some complex solution of the problem, but I don't know how to solve complex integration. What I did is :

$$\int \frac{1}{x} \int \left(\frac {1}{\sqrt {1-x^2}} \, dx\right) \, dx$$

$$= \int y \cot y \, dy \text{ where } y = \arcsin x$$

$$= y \ln |\sin y| - \int \ln |\sin y |dy$$

So will you plz help me in one of the two ways:

1. $\displaystyle\int y \cot y \, dy = \text{?}$ or $\displaystyle\int \ln |\sin y | \, dy = \text{?}$

2. Any better/neater solution of the original problem ?

I saw this question in MSE but in definite integral form, (which is okay), but not sure about this indefinite form.Thanks.

Answer 1

I do not think that we could find a closed form expression except using very special and complex function.

If $$I=\int \frac {\log (x)}{\sqrt {1-x^2}}dx$$ just as you did using integration by parts $$u=\log(x)\implies du=\frac{dx}x$$ $$dv=\frac {dx}{\sqrt {1-x^2}}\implies v=\sin^{-1}(x)$$ then $$I=\log (x) \sin ^{-1}(x)-\int \frac{\sin ^{-1}(x)}{x}\,dx$$ This last integral could be computed using Taylor series $$\sin^{-1}(x)=\sum^{\infty}_{n=0} \frac{(2n)!}{4^n (n!)^2 (2n+1)} x^{2n+1}\quad\text{ for }|x| \le 1$$ which makes, integrting each term of the expansion, $$\int \frac{\sin ^{-1}(x)}{x}\,dx=\sum^{\infty}_{n=0} \frac{(2n)!}{4^n (n!)^2 (2n+1)^2} x^{2n+1}$$

Just for your curiosity, this last summation corresponds to $$\int \frac{\sin ^{-1}(x)}{x}\,dx=x \,\, _3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2};x^2\right )$$ where appears the generalized hypergeometric function.

What is amazing (at least to me !) is that this monster is quite "close" to $\frac{\pi \log (2)}{2}x $.

Edit

Another possible way is to start with the expansion $$\frac {1}{\sqrt {1-x^2}}=\sum_{n=0}^ \infty \frac{\binom{2 n}{n} }{4^n}x^{2n}$$ That is to say that $$I=\int \frac {\log (x)}{\sqrt {1-x^2}}=\sum_{n=0}^ \infty \frac{\binom{2 n}{n} }{4^n\,(2n+1)^2}\left((2n+1)\log(x)-1 \right)x^{2n+1}$$ $$I=x (\log (x)-1) \, _3F_2\left(\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2};x^2\right )+\frac{1}{9} x^3 \log (x) \, _3F_2\left(\frac{3}{2},\frac{3}{2},\frac{3}{2};\frac{5}{2},\frac{5}{2};x^2\right )$$

Answer 2

One does not need hypergeometric functions to find this antiderivative. It amounts to use poly-logarithms only. We have: \begin{eqnarray} \int \frac{\log(x)}{\sqrt{1-x^2}} dx &\underbrace{=}_{x=\sin(u)}& \int \log(\sin(u) du\\ &\underbrace{=}_{t=\exp(\imath u)}& \int \log(\frac{t-1/t}{2 \imath})\cdot \frac{dt}{\imath t}\\ &=& \int \frac{\log(t-1)+\log(t+1)-\log(2 \imath t)}{\imath t} dt\\ &=& \frac{1}{2} i \left(2 \text{Li}_2(-t)+2 \text{Li}_2(t)+\log ^2(2 i t)+2 \log (1-t) \log (t)-2 \log (t-1) \log (t)\right)\\ &=& \frac{1}{2} i \left(2 \text{Li}_2\left(-i x-\sqrt{1-x^2}\right)+2 \text{Li}_2\left(i x+\sqrt{1-x^2}\right)+2 \pi \arcsin(x)+\left(i \arcsin(x)+\frac{i \pi }{2}+\log (2)\right)^2\right) \end{eqnarray} In particular we have: \begin{equation} \int\limits_0^1 \frac{\log(x)}{\sqrt{1-x^2}} dx = -\frac{\pi}{2} \log(2) \end{equation}