$$ I = \frac{\pi^2}{8} – \int_0^1 \frac{\arctan(x)}{\sqrt{1-x^2}} \,dx $$
Evaluate $I$
$$ I = \frac{\pi^2}{8} – \int_0^1 \frac{\arctan(x)}{\sqrt{1-x^2}} \,dx$$
$$f(a) = \int_0^1 \frac{\arctan(ax)}{\sqrt{1-x^2}} \,dx$$
After differentiating and integrating,
$$ f'(a) = \frac{\ln(a+\sqrt{1+a^2})}{a\sqrt{1+a^2}} $$
The final answer that we get is: $$ f(1) = \frac{\pi^2}{8} – \frac{\ln^2(1+\sqrt{2})}{2} $$
Could somebody please show me how this was done? $$$$ Also, how do we know how and where to introduce another variable (say $n$) to perform Feynman Integration? For example, in this case, how do we know that we should rewrite $$f(x) = \int_0^1 \frac{\arctan(x)}{\sqrt{1-x^2}} \,dx$$ as $$f(x,a) = \int_0^1 \frac{\arctan(ax)}{\sqrt{1-x^2}} \,dx$$ $$$$Please do help me!
Best Answer
There is no general rule in introducing an extra variable to apply the Feynman differentiation trick. In this case it looks reasonable to consider $\arctan(x)$ as $\arctan(ax)|_{a=1}$ because the derivative of the arctangent is a rather simple algebraic function. So, let: $$ I(a) = \int_{0}^{1}\frac{\arctan(ax)}{\sqrt{1-x^2}}\,dx. \tag{1}$$ We trivially have $I(0)=0$ and: $$ \frac{d}{da}\,I(a) = \int_{0}^{1}\frac{x}{(1+x^2 a^2)\sqrt{1-x^2}} = \int_{0}^{\pi/2}\frac{\sin\theta}{1+a^2\sin^2\theta}\,d\theta \tag{2}$$ is an integral that can be solved through the Weierstrass substitution: $$ \int_{0}^{\pi/2}\frac{\sin\theta}{1+a^2\sin^2\theta}\,d\theta = 4\int_{0}^{1}\frac{t}{(1+t^2)^2+4a^2 t^2}\,dt =\frac{1}{a\sqrt{1+a^2}}\,\text{arctanh}\frac{a}{\sqrt{1+a^2}}\tag{3}$$ leading to: $$ \int_{0}^{1}\frac{\arctan(x)}{\sqrt{1-x^2}}\,dx=\int_{0}^{1}\frac{1}{a\sqrt{1+a^2}}\,\text{arctanh}\frac{a}{\sqrt{1+a^2}}\,da\tag{4}$$ and by replacing $a$ with $\sinh u$ we get: $$\begin{eqnarray*} \int_{0}^{1}\frac{\arctan(x)}{\sqrt{1-x^2}}\,dx &=& \int_{0}^{\text{arcsinh}(1)}\frac{u\,du}{\sinh u}\\&=&u\cdot\left.\log\tanh\frac{u}{2}\right|_{0}^{\text{arcsinh}(1)}-\int_{0}^{\text{arcsinh}(1)}\log\tanh\frac{u}{2}\,du\\&=&-\log^2(1+\sqrt{2})+\int_{0}^{\log(1+\sqrt{2})}\log\coth(u/2)\,du\tag{5}\end{eqnarray*}$$ and we have just to check that: $$\int\log\coth u\,du = \frac{1}{2}\left(\log(\coth u)\cdot\log(1+\coth u)+\text{Li}_2(1-\coth u)+\text{Li}_2(-\coth u)\right)\tag{6}$$ then use the usual functional equations for the dilogarithm.