Solving integrals through “integral systems.”

Question

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:

$$I=-\frac{\pi}{4}\log(2)-\frac{1}{2}G$$

$$J= -\frac{\pi}{4}\log(2)+\frac{1}{2}G$$

Answer 1

A related, yet more advanced, system of integrals: $$I=\int_{0}^{\frac{\pi}{4}} y \ln (2\sin y) d y,\>\>\> J=\int_{0}^{\frac{\pi}{4}} y \ln (2\cos y) d y $$ along with $K=\int_{0}^{\frac{\pi}{2}} y \ln (\tan y) d y $. \begin{align} &I+J=\int_{0}^{\frac{\pi}{4}} y \ln \overset{2y\to y}{(2\sin 2y) }d y =\frac14 \int_{0}^{\frac{\pi}{2}} \overset{y\to \frac\pi2-y} {y \ln (2\sin y)}{ d y}= \frac18K\\ &I-J =\int_{0}^{\frac{\pi}{4}} y \ln (\tan y) d y = K - \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \overset{y\to \frac\pi2-y} {y \ln (\tan y)}{d y}=\frac12K-\frac\pi4G\\ &K \overset{t=\tan y}=\int_{0}^{\infty} \frac{\ln t\tan^{-1}t}{1+t^2}dt = \int_{0}^{1} \int_0^\infty \frac{t\ln t}{(1+t^2 )(1+s^2 t^2)} \overset{t\to \frac1{st}}{dt}ds \\ &=\frac12\int_{0}^{1} \int_0^\infty \frac{-t\ln s}{(1+t^2 )(1+s^2 t^2)} dt\>ds= \frac12 \int_0^1\frac{\ln^2s}{1-s^2}ds=\frac78\zeta(3) \end{align} Solve to obtain $$I= -\frac\pi8G +\frac{35}{128}\zeta(3), \>\>\>\>\>J= \frac\pi8G -\frac{21}{128}\zeta(3) $$