Hello fellow MSE users I have recently evaluated two integrals simultaneously and was wondering if any of you had other examples of solving integrals this way, specifically non-elementary definite integrals please! Here is my work as to illustrate the technique I am talking about.
Consider the following two integrals:
$$I:=\int_{0}^{\frac{\pi}{4}}\log(\sin(x))\,dx$$
$$J:=\int_{0}^{\frac{\pi}{4}}\log(\cos(x))\,dx$$
So it follows that
$$I+J=\int_{0}^{\frac{\pi}{4}}\log(\frac{1}{2}\sin(2x))\,dx$$
$$=-\frac{\pi}{4}\log(2)+\int_{0}^{\frac{\pi}{4}}\log(\sin(2x))\,dx$$
Let $$2x\longrightarrow{x}$$
$$=-\frac{\pi}{4}\log(2)+\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\log(\sin(x))\,dx$$
The integral left is one of Euler’s famous integrals which has a well known value. So implementing this and doing the arithmetic it follows that:
$$I+J=-\frac{\pi}{2}\log(2)$$
Then we consider
$$I-J=\int_{0}^{\frac{\pi}{4}}\log(\tan(x))\,dx$$
Let $$\tan(x)\longrightarrow{x}$$
$$=\int_{0}^{1}\frac{\log(x)}{x^2+1}\,dx$$
Using a series expansion it is elementary to prove that
$$I-J=-G$$
Where G is Catalan’s constant
So we have the following 2x2 linear system, namely:
$$I+J=-\frac{\pi}{2}\log(2)$$ $$I-J=-G$$
Solving it yields:
$$I=-\frac{\pi}{4}\log(2)-\frac{1}{2}G$$
$$J= -\frac{\pi}{4}\log(2)+\frac{1}{2}G$$
