# Prove $\int_0^2 \frac{\log{x}}{\sqrt{4-x^2}}\text{d}x=0$ without integrating

Inspired by this post:

Prove that $\int_0^4 \frac{\ln x}{\sqrt{4x-x^2}}~dx=0$ (without trigonometric substitution)

I was wondering if there exists a similar pair of substitutions for the following integral:

$$I = \int_0^2 \frac{\log{x}}{\sqrt{4-x^2}}\text{d}x$$

More precisely:

Does there exist a linear and non-linear transformation (or possibly two non-linear transformations) $$u(x), v(x)$$ such that by applying them to the above integral, we can obtain $$aI = bI$$ where $$\mathbb{R} \ni a \neq b \in \mathbb{R}$$

I have clearly overthought this.

Substitute $$x \mapsto x^2$$ to transform $$I$$ into a quarter of the integral linked.

Nesting the transformations, we find the appropriate non-linear transformations for this integral are:

$$u\mapsto \sqrt{4-x^2}, u \mapsto x^2 - 2$$

Resulting in:

$$I=\frac{1}{2}\int_0^2\frac{\log{(4-u^2)}}{\sqrt{4-u^2}}\text{d}u$$

$$I=\frac{1}{4}\int_{-2}^2\frac{\log{(2+u)}}{\sqrt{4-u^2}}\text{d}u=\frac{1}{4}\int_0^2\frac{\log{(4-u^2)}}{\sqrt{4-u^2}}\text{d}u$$

$$I = \frac{\mathcal{I}}{2} = \frac{\mathcal{I}}{4}$$

• Nice approach (+1) Commented Mar 27, 2019 at 11:32
• Nice gymnastics ! and $\to +1$ for sure. Commented Mar 27, 2019 at 11:34

I found a way of proving that $$I=2I$$ without integrating, although it is not similar to the one explained in the linked post.

Let $$x=2\sin(t)$$ then \begin{align}I&=\int_0^{\pi/2}\log(2\sin(t))\,dt\\ &=\int_0^{\pi/2}\log(4\sin(t/2)\cos(t/2))\,dt\\ &=\int_0^{\pi/2}\log(2\sin(t/2))\,dt+\int_0^{\pi/2}\log(2\cos(t/2))\,dt\\ &=\int_0^{\pi/2}\log(2\sin(t/2))\,dt+\int_{\pi/2}^{\pi}\log(2\sin(s/2))\,ds\\ &=\int_0^{\pi}\log(2\sin(s/2))\,ds=2\int_0^{\pi/2}\log(2\sin(u))\,du=2I \end{align} where $$s=\pi-t$$ and $$u=s/2$$. Hence $$I=0$$.

• While this is a quite neat way of proving the given equality, the OP asks for a similar pair of substitutions $($similiar to the linked post$)$, namely a pair of linear and non-linear transformation in order to achieve the result. Commented Mar 27, 2019 at 9:22
• I missed the day when you reached the $100k$ reputation. Congratulations and thanks for posting so good answers so often. Cheers :-) Commented Mar 27, 2019 at 9:38
• @ClaudeLeibovici Thanks for your generous comments! Commented Mar 27, 2019 at 11:26
• @mrtaurho It seems that OP found the appropriate non-linear transformations Commented Mar 27, 2019 at 11:27