Introduce a parameter to determine the value of $\int_0^1\frac{\log(1+x)}{1+x^2}dx$

Question

How could I introduce a parameter to determine the value of $\int_0^1\frac{\log(1+x)}{1+x^2}dx$?

Answer 1

You can write the integral as $$I(\alpha)=\int_0^1 \frac{\log(1+\alpha x)}{1+x^2}\,dx$$
The target integral is the case $\alpha=1$, note that for $\alpha=0$ we find $I(0)=0$ which is going to be useful later. Now we want to differentiate $I(\alpha)$ with respect to $\alpha$, so we write: $$I'(\alpha)=\frac{\partial}{\partial\alpha}I(\alpha)=\frac{\partial}{\partial\alpha}\int_0^1 \frac{\log(1+\alpha x)}{1+x^2}\,dx=\int_0^1 \frac{\partial}{\partial\alpha}\frac{\log(1+\alpha x)}{1+x^2}\,dx$$ We can bring inside the integral the partial differential because of the theorem of Dominated Convergence. $$I'(\alpha)=\int_0^1 \frac{x}{(1+x^2)(1+\alpha x)}\,dx=\int_0^1\frac{\alpha+x}{(a+x^2)(a+\alpha x)}-\frac{\alpha}{(\alpha^2+1)(1+(\alpha x))}\,dx$$ Integrating with respect of $x$ we obtain an expression that is in $\alpha$: $$I'(\alpha)=\frac{\pi}{4}\frac{\alpha}{\alpha^2+1}-\frac{\log{2}}{2(\alpha^2+1)}$$ We now recover the original function by integrating this derivative: $$I(\alpha)=\int \frac{\pi}{4}\frac{\alpha}{\alpha^2+1}-\frac{\log{2}}{2(\alpha^2+1)}\,d\alpha=\frac{\pi}{8}\log{\alpha^2+1}-\frac{\log{2}\arctan(\alpha^2+1)}{2} +C$$ By chosing $\alpha = 0$ we recover that $C=0$ as well.
We end up with: $$\int_0^1 \frac{\log(1+x)}{1+x^2}\,dx=I(1)=\frac{\pi}{8}\log{2}-\frac{\log{2}\arctan(2)}{2}$$