How can I evaluate $\int _0^1\frac{\tan ^{-1}\left(3\sqrt{\frac{a}{4-a}}\right)}{\sqrt{a}\sqrt{4-a}}\:\mathrm{d}a$

Question

I have the integral: $$\int _0^1\frac{\tan ^{-1}\left(3\sqrt{\frac{a}{4-a}}\right)}{\sqrt{a}\sqrt{4-a}}\:\mathrm{d}a.$$ If I use $u=3\sqrt{\frac{a}{4-a}}$, I get $$6\int _0^{\sqrt{3}}\:\frac{\tan ^{-1}\left(u\right)}{u^2+9}\mathrm{d}u$$ But I cant get past this. I tried integration by parts but nothing seems to come out from that.

Answer 1

Using $$ \tan^{-1}(x) = \frac{i}{2} \, \ln\left( \frac{1 - i \, x}{1 + i \, x} \right) $$ then \begin{align} \int \frac{\tan^{-1}(x)}{x^2 + a^2} \, dx &= \frac{i}{2} \, \left( \int \frac{\ln(1 - i \, x)}{x^2 + a^2} \, dx - \int \frac{\ln(1 + i \, x)}{x^2 + a^2} \, dx \right). \end{align} Using \begin{align} \int \frac{\ln(1 + i \, x)}{x^2 + a^2} \, dx &= \frac{i}{2 \, a} \, \left[ - \text{Li}_{2}\left(\frac{1+ i \, x}{1 - a}\right) + \text{Li}_{2}\left( \frac{1 + i \, x}{1 + a}\right) + \ln(1 + i \, x) \, \ln\left( \frac{a-1}{a+1} \cdot \frac{a - i \, x}{a + i \, x}\right) \right] \\ \int \frac{\ln(1 - i \, x)}{x^2 + a^2} \, dx &= \frac{i}{2 \, a} \, \left[ - \text{Li}_{2}\left(\frac{1- i \, x}{1 + a}\right) + \text{Li}_{2}\left( \frac{1 - i \, x}{1 - a}\right) + \ln(1 - i \, x) \, \ln\left( \frac{a+1}{a-1} \cdot \frac{a - i \, x}{a + i \, x}\right) \right] \end{align} then the result can be found in closed form with: \begin{align} \int \frac{\tan^{-1}(x)}{x^2 + a^2} \, dx &= \frac{-1}{4 \, a} \, \left[ - \text{Li}_{2}\left(\frac{1- i \, x}{1 + a}\right) - \text{Li}_{2}\left(\frac{1 + i \, x}{1 + a}\right) + \text{Li}_{2}\left(\frac{1 - i \, x}{1 - a}\right) + \text{Li}_{2}\left(\frac{1 + i \, x}{1 - a}\right) \\+ \ln(1 + x^2) \, \ln\left(\frac{a+1}{a-1}\right) - 4 \, \tan^{-1}(x) \, \tan^{-1}\left(\frac{x}{a}\right) \right] \end{align}

The integral with the desired limits is: $$ \int_{0}^{\sqrt{3}} \frac{\tan^{-1}(x)}{x^2 + 3^2} \, dx = \frac{1}{6} \, \left( \frac{\zeta(2)}{3} - \frac{\ln(2) \, \ln(3)}{2} - \frac{\text{Li}_{2}(-3)}{4} \right). $$

Also: $$ \int_{0}^{1} \tan^{-1}\left( 3 \, \sqrt{\frac{x}{4-x}} \right) \, \frac{dx}{\sqrt{x \, (4-x)}} = \frac{\zeta(2)}{3} - \frac{\ln(2) \, \ln(3)}{2} - \frac{\text{Li}_{2}(-3)}{4}. $$

Answer 2

For the antiderivative, we could combine two things $$ \frac 6 {u^2+9}=\frac{i}{u+3 i}-\frac{i}{u-3 i}$$ $$\tan^{-1}(u) = \frac{i}{2} \, \log(1-iu)-\frac{i}{2} \, \log(1+iu) $$ and make $$I=6\int\frac{\tan ^{-1}\left(u\right)}{u^2+9}\,du=$$ $$\frac 12 \left( \frac{\log (1-i u)}{u-3 i}-\frac{\log (1-i u)}{u+3 i}-\frac{\log (1+i u)}{u-3 i}+\frac{\log (1+i u)}{u+3 i}\right)$$ and recall that $$\int \frac{\log (1+i u)}{u+i b}\,du=\text{Li}_2\left(\frac{1+i u}{1+b}\right)+\log (1+i u) \log \left(\frac{b-iu}{1+b}\right)$$ $$\int \frac{\log (1-i u)}{u+i b}\,du=\text{Li}_2\left(\frac{1-i u}{1-b}\right)+\log (1-i u) \log \left(\frac{-b+i u}{1-b}\right)$$ and the definite integral is $$\frac{\pi ^2}{12}-\frac{\log ^2(2)}{2}+\frac 12 \left(\text{Li}_2\left(\frac{1}{4} \left(1-i \sqrt{3}\right)\right)+\text{Li}_2\left(\frac{1}{4} \left(1+i \sqrt{3}\right)\right)-\text{Li}_2\left(\frac{1}{4}\right) \right)$$ which, once simplified, gives the result already provided in comments and answers.