# Evaluate $\int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)$

I am trying to evaluate the following integral $$I(k) = \int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)$$ with $0< k < 1$.

My attempt

By performing the substitution $$y=\frac{1-x}{1+x} \Longleftrightarrow x=\frac{1-y}{1+y}$$

we have $$I(k) = \int_0^1 \mathrm{d}y \frac{\log y}{\sqrt{y}}\frac{y-1}{(k^2-1)y^2-2y(k^2+1)+(k^2-1)}$$

Now we can decompose $$\frac{y-1}{(k^2-1)y^2-2y(k^2+1)+(k^2-1)}= \frac{1}{2k(1+k)}\frac{1}{y+a}- \frac{1}{2k(1-k)}\frac{1}{y+a^{-1}}$$

with $a=\frac{1-k}{1+k}$ runs from $0$ to $1$. Therefore we can write $$I(k) = \frac{1}{2k(1+k)}\color{blue}{\int_0^1 \mathrm{d}y \frac{\log y}{\sqrt{y}}\frac{1}{y+a}} - \frac{1}{2k(1-k)}\color{red}{\int_0^1 \mathrm{d}y \frac{\log y}{\sqrt{y}}\frac{1}{y+a^{-1}}}$$

So we have only to evaluate $\color{blue}{I_1}$ and $\color{red}{I_2}$.

Now we consider the following integral $$J(\sigma) = \int_0^1 \mathrm{d}x \frac{\log x}{\sqrt{x}}\frac{1}{x-\sigma^2}$$ so that $\color{blue}{I_1}=J(i\sqrt{a})$ and $\color{red}{I_2}=J(i/\sqrt{a})$.

By considering the map $x\mapsto x^2$ we can write $$J(\sigma)=4\int_0^1\mathrm{d}x \frac{\log x}{x^2-\sigma^2}= \frac{2}{\sigma}\left[\color{green}{\int_0^1 \frac{\log x}{x-\sigma}}-\color{green}{\int_0^1 \frac{\log x}{x+\sigma}}\right]$$

The problem then reduces to evaluate the $\color{green}{\text{green}}$ integrals. At this point I'm stuck. I think that it needs to be solved by using polylogarithms, but I don't really know how to use these functions.

Mathematica 11.0 says

$$J(\sigma)=4 \left(\frac{\Phi \left(\frac{1}{\sigma ^2},2,\frac{3}{2}\right)}{4 \sigma ^4}+\frac{1}{\sigma ^2}\right)$$

where $\Phi$ is the Lerch transcendent. I don't know if this result is true (numerical integration is somewhat problematic). However, if it is true, I don't know what to do next.

Any hint on how to proceed with the evaluation?

Another method which gives a closed form result. Using the identity $$\ln\frac{1-x}{1+x}=-2\operatorname{arctanh} x$$ and parity of the integrand, the integral can be written as \begin{align} I(k) &= \int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)\\ &=-\int_{-1}^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\operatorname{arctanh} x\\ &=-\int_{-\infty}^\infty \frac{u\sinh u}{\cosh^2u-k^2\sinh^2u}du \end{align} The last integral is obtained by the substitution $x=\tanh u$. Then \begin{align} I(k)&=-\frac{1}{2k}\left[J(k)-J(-k)\right]\\ J(k)&=\int_{-\infty}^\infty \frac{udu}{\cosh u-k\sinh u} \end{align}

To evaluate $J(k)$, we enforce the substitution $t=e^u$ and, denoting $m^2=\frac{1+k}{1-k}$, we obtain \begin{align} J(k)&=\frac{2}{1-k}\int_0^\infty \frac{\ln v}{v^2+m^2}dv\\ &=\frac{2}{m(1-k)}\int_0^\infty \frac{\ln mt}{t^2+1}dt\\ &=\frac{2}{m(1-k)}\left[\int_0^\infty \frac{\ln t}{t^2+1}dt+\int_0^\infty \frac{\ln m}{t^2+1}dt\right] \end{align} We know that $\int_0^\infty \frac{\ln t}{t^2+1}dt=0$, thus $$J(k)=\frac{\pi \ln m}{m(1-k)}=\frac{\pi}{2}\frac{\ln \frac{1+k}{1-k}}{\sqrt{1-k^2}}$$ Finally, $$I(k)=\frac{\pi}{2k\sqrt{1-k^2}}\ln\frac{1- k}{1+ k}$$

• That's really amazing! I will update my question with the other method. Commented Feb 5, 2018 at 10:02

A great approach on your part! Let me try another.

$$I(k) = \int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)=-2 \sum_{n=0}^\infty \frac{1}{2n+1} \int_0^1 \frac{x^{2n+2}}{\sqrt{1-x^2}}\frac{\mathrm{d}x }{1-k^2x^2}$$

So we really only need to solve the integral in the form:

$$Y(k,n+1)=\int_0^1 \frac{x^{2n+2}}{\sqrt{1-x^2}}\frac{\mathrm{d}x }{1-k^2x^2}$$

Note that:

$$Y(k,0)=\int_0^1 \frac{\mathrm{d}x }{\sqrt{1-x^2}(1-k^2x^2)}=\frac{\pi}{2 \sqrt{1-k^2}}$$

This can be seen directly from the antiderivative (check by differentiation):

$$\int \frac{\mathrm{d}x }{\sqrt{1-x^2}(1-k^2x^2)} =\frac{1}{\sqrt{1-k^2}} \tan^{-1} \left( \frac{\sqrt{1-k^2} x}{\sqrt{1-x^2}} \right)$$

Let's try something for the general case:

$$Y(k,n+1)=\frac{1}{k^2}\int_0^1 \frac{x^{2n}(k^2x^2-1+1)}{\sqrt{1-x^2}}\frac{\mathrm{d}x }{1-k^2x^2}=$$

$$=-\frac{1}{k^2} \int_0^1 \frac{x^{2n} \mathrm{d}x}{\sqrt{1-x^2}}+\frac{1}{k^2} \int_0^1 \frac{x^{2n}}{\sqrt{1-x^2}}\frac{\mathrm{d}x }{1-k^2x^2}$$

From which we write a recurrence relation:

$$Y(k,n+1)=-\frac{1}{k^2} \int_0^1 \frac{x^{2n} \mathrm{d}x}{\sqrt{1-x^2}}+\frac{1}{k^2} Y(k,n)$$

Let us denote:

$$P(n)=\int_0^1 \frac{x^{2n} \mathrm{d}x}{\sqrt{1-x^2}}$$

Then:

$$Y(k,n+1)=-\frac{1}{k^2} \left( P(n)-Y(k,n) \right)$$

$P(n)$ is just Beta function and has the form:

$$P(n)=\frac{\sqrt{\pi}~ \Gamma \left(n+\frac{1}{2} \right)}{2 n!}$$

In other words, while this solution is by no means complete, we already obtained:

The series representation of the integral:

$$I(k)=-2 \sum_{n=0}^\infty \frac{1}{2n+1} Y(k,n+1) \tag{1}$$

The recurrence relation together with the initial condition, which allows us to compute all the terms:

$$Y(k,n+1)=\frac{1}{k^2} \left(Y(k,n)- \frac{\sqrt{\pi}~ \Gamma \left(n+\frac{1}{2} \right)}{2 n!} \right) \tag{2}$$

$$Y(k,0)=\frac{\pi}{2 \sqrt{1-k^2}} \tag{3}$$

I will check how this works numerically in a couple of days.

As can be seen from the integral, $Y(k,n)$ actually has a solution in terms of Gauss Hypergeometric function:

$$Y(k,n)=\frac{\sqrt{\pi}~ \Gamma \left(n+\frac{1}{2} \right)}{2 n!} {_2 F_1} \left(1, n+\frac{1}{2};n+1; k^2 \right)$$

But the recurrence relation is a more simple way to obtain the values without special software. $P(n)$ can be converted into the form of rational multiples of $\pi$.

Final remark: while this doesn't seem to lead us to a closed form, let me note that this approach works for a large class of integrals, as long as we can expand the non-algebraic function as a series.

Trying to derive a rational form for $P(n)$:

$$\Gamma \left(n+\frac{1}{2} \right)=\left(n-1+\frac{1}{2} \right)\left(n-2+\frac{1}{2} \right) \cdots \left(1+\frac{1}{2} \right) \frac{1}{2} \Gamma \left(\frac{1}{2} \right)=$$

$$=\left(n-1+\frac{1}{2} \right)\left(n-2+\frac{1}{2} \right) \cdots \left(1+\frac{1}{2} \right) \frac{\sqrt{\pi}}{2}=$$

$$=\left(2n-1 \right)\left(2n-3 \right) \cdots 3 \cdot \frac{\sqrt{\pi}}{2^n}$$

So:

$$P(n)=\frac{\pi}{2} \frac{1 \cdot 3 \cdots (2n-3)(2n-1) }{2^n~n!}$$

Separate case: $P(0)= \pi/2$.

Edit 2

I have checked the algorithm numerically, and it works, but the convergence is terrible, and a huge loss of significance is involved.

• Thanks Yuriy!! That's a great approach too! However, I accept the answer from Jack, since it is more useful for the way I've started attacking the problem. Commented Feb 5, 2018 at 0:35
• Of course, I completely understand Commented Feb 5, 2018 at 0:37

Nice approach so far. About the green integrals, you may simply expand $\frac{1}{x\pm\sigma}$ as a geometric series and exploit $\int_{0}^{1} x^{m-1}\log(x)\,dx =-\frac{1}{m^2}$ to get $$\int_{0}^{1}\frac{-\log x}{\sigma-x}\,dx =\frac{1}{\sigma}\sum_{m\geq 1}\frac{1}{\sigma^m m^2}=\frac{1}{\sigma}\,\text{Li}_2\left(\frac{1}{\sigma}\right)$$ and similarly $$\int_{0}^{1}\frac{-\log x}{\sigma+x}\,dx =\frac{1}{\sigma}\,\text{Li}_2\left(-\frac{1}{\sigma}\right).$$ Then the reflection formulas for the dilogarithm might induce some extra simplification.
They did it here, for instance.

• Thanks Jack! I will try to evaluate the integrals by doing this. I will update the question when I have a more complete answer. Commented Feb 5, 2018 at 0:40

I have completed the computation of ASM.

For $0<k<1$,

$\displaystyle I(k) = \int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)\,dx$

Let $\rho=\dfrac{1-k}{1+k}$

Observe that, since $0<k<1$ then $0<\rho<1$.

Perform the change of variable $y=\dfrac{1-x}{1+x}$,

\begin{align}I(k)&=\int_0^1 \frac{(y-1)\ln y}{\sqrt{y}\big(\left(k-1\right)y-k-1\big)\big(\left(k+1\right)y-k+1\big)}\,dy\\ &=\frac{1}{(k^2-1)}\int_0^1 \frac{(y-1)\ln y}{\sqrt{y}\left(y+\frac{1}{\rho}\right)\left(y+\rho\right)}\,dy \end{align}

Perform the change of variable $\displaystyle t=\sqrt{y}$,

\begin{align}I(k)&=\frac{4}{(k^2-1)}\int_0^1 \frac{(t^2-1)\ln t}{\left(t^2+\frac{1}{\rho}\right)\left(t^2+\rho\right)}\,dt\\ &=\frac{4\rho}{(\rho-1)(k^2-1)}\left(\int_0^1 \frac{\ln t}{t^2+\rho}\,dt-\int_0^1 \frac{\ln t}{\rho t^2+1}\,dt\right)\\ \end{align}

In the latter integral perform the change of variable $\displaystyle u=\frac{1}{t}$,

\begin{align}I(k)&=\frac{4\rho}{(\rho-1)(k^2-1)}\left(\int_0^1 \frac{\ln t}{t^2+\rho}\,dt+\int_1^{\infty} \frac{\ln u}{u^2+\rho}\,du\right)\\ &=\frac{4\rho}{(\rho-1)(k^2-1)}\int_0^{\infty} \frac{\ln t}{t^2+\rho}\,dt\\ \end{align}

Since $\displaystyle \rho>0$,

\begin{align}I(k)&=\frac{4}{(\rho-1)(k^2-1)}\int_0^{\infty} \frac{\ln t}{\left(\frac{t}{\sqrt{\rho}}\right)^2+1}\,dt\end{align}

Perform the change of variable $v=\dfrac{t}{\sqrt{\rho}}$,

\begin{align}I(k)&=\frac{4\sqrt{\rho}}{(\rho-1)(k^2-1)}\int_0^{\infty} \frac{\ln\left( v\sqrt{\rho}\right)}{v^2+1}\,dv\\ &=\frac{2\sqrt{\rho}\ln\left(\rho\right)}{(\rho-1)(k^2-1)}\int_0^{\infty} \frac{1}{v^2+1}\,dv+\frac{4\sqrt{\rho}}{(\rho-1)(k^2-1)}\int_0^{\infty} \frac{\ln\left( v\right)}{v^2+1}\,dv\\ &=\frac{2\sqrt{\rho}\ln\left(\rho\right)}{(\rho-1)(k^2-1)}\Big[\arctan v\Big]_0^{\infty}+\frac{4\sqrt{\rho}}{(\rho-1)(k^2-1)}\int_0^{\infty} \frac{\ln\left( v\right)}{v^2+1}\,dv\\ &=\frac{\pi\sqrt{\rho}\ln\left(\rho\right)}{(\rho-1)(k^2-1)}+\frac{4\sqrt{\rho}}{(\rho-1)(k^2-1)}\int_0^{\infty} \frac{\ln\left( v\right)}{v^2+1}\,dv\\ \end{align}

But,

Consider the integral,

$\displaystyle J=\int_0^{\infty} \frac{\ln\left( v\right)}{v^2+1}\,dv$

Perform the change of variable $w=\dfrac{1}{v}$,

$\displaystyle J=-J$.

Therefore $\displaystyle J=0$,

and,

\begin{align}J(k)&=\frac{\pi\sqrt{\rho}\ln\left(\rho\right)}{(\rho-1)(k^2-1)}\\ &=\frac{\pi\sqrt{\frac{1-k}{1+k}}\ln\left(\frac{1-k}{1+k}\right)}{(\frac{1-k}{1+k}-1)(k^2-1)}\\ &=\frac{\pi\sqrt{\frac{1-k}{1+k}}\ln\left(\frac{1-k}{1+k}\right)}{(\frac{-2k}{1+k})(k^2-1)}\\ &=\frac{\pi\sqrt{\frac{1-k}{1+k}}\ln\left(\frac{1-k}{1+k}\right)}{2k(1-k)}\\ &=\boxed{\frac{\pi\ln\left(\frac{1-k}{1+k}\right)}{2k\sqrt{1-k^2}}} \end{align}