# Evaluate $\int_0^4 \frac{\ln x}{\sqrt{4x-x^2}} \,\mathrm dx$

Evaluate
$$\displaystyle\int_0^4 \frac{\ln x}{\sqrt{4x-x^2}} \,\mathrm dx$$

How do I evaluate this integral? I know that the result is $0$, but I don’t know how to obtain this. Wolfram|Alpha yields a non-elementary antiderivative for the indefinite integral, so I don’t think I can directly integrate and then plug in the upper/lower limits.

#### Solutions Collecting From Web of "Evaluate $\int_0^4 \frac{\ln x}{\sqrt{4x-x^2}} \,\mathrm dx$"

First let $t = x-2$ this way $4x-x^2 = 4 – (x-2)^2 = 4-t^2$. Substitute,
$$\int_{-2}^2 \frac{\log(t+2)}{\sqrt{4-t^2}} ~ dt$$
Now let, $\theta = \sin^{-1}\tfrac{t}{2}$ so that $2\sin \theta = t$ and hence, after substitute,
$$\int_{-\pi/2}^{\pi/2} \frac{\log [2(1+\sin \theta)]}{2\cos \theta} 2\cos \theta ~ d\theta = \pi \log 2 + \int_{-\pi/2}^{\pi/2} \log(1+\sin \theta)~d\theta$$
To solve this integral, replace $\theta$ by $-\theta$,
$$I = \int_{-\pi/2}^{\pi/2} \log(1+\sin \theta) ~d\theta= \int_{-\pi/2}^{\pi/2} \log(1-\sin \theta)~d\theta$$
Now,
$$I + I = \int_{-\pi/2}^{\pi/2} \log(1-\sin^2 \theta) ~ d\theta = 4\int_{0}^{\pi/2} \log (\cos \theta) ~ d\theta$$
The last integral is a well-known integral that computes to $-\frac{\pi}{2}\log 2$.

Your final answer is, $\pi \log 2 -\pi\log 2$.

This integral appeared on an 1886 exam at the University of Cambridge and also discussed in A Treatise on the Integral Calculus by Joseph Edwards. In general we have

$$\int_0^a \frac{\ln x}{\sqrt{ax-x^2}}\,\mathrm dx=\pi\ln\left(\frac{a}{4}\right)$$

Proof :

\begin{align}
\int_0^a \frac{\ln x}{\sqrt{ax-x^2}}\,\mathrm dx&=\int_0^1 \frac{\ln a+\ln t}{\sqrt{t}\;\sqrt{1-t}}\,\mathrm dt\tag1\\[7pt]
&=\int_0^{\pi/2}\frac{\ln a+\ln\sin^2\theta }{\sqrt{\sin^2\theta}\;\sqrt{1-\sin^2\theta}}\cdot2\sin\theta\cos\theta\;\mathrm d\theta\tag2\\[7pt]
&=2\ln a\int_0^{\pi/2}\;\mathrm d\theta+4\int_0^{\pi/2}\ln\sin\theta\;\mathrm d\theta\tag3\\[7pt]
&=\pi\ln a-2\pi\ln2\\[7pt]
&=\bbox[5pt,border:3px #FF69B4 solid]{\color{red}{\large\pi\ln\left(\frac{a}{4}\right)}}\tag{$\color{red}{❤}$}
\end{align}
Hence
$$\int_0^4 \frac{\ln x}{\sqrt{4x-x^2}}\,\mathrm dx=\bbox[5pt,border:3px #FF69B4 solid]{\color{red}{\large0}}$$

Explanation :

$(1)\;$ Use substitution $\;\displaystyle x=at$

$(2)\;$ Use substitution $\;\displaystyle t=\sin^2\theta\quad\implies\quad dt=2\sin\theta\cos\theta\;\mathrm d\theta$

$(3)\;$ Use Euler log-sine integral $\;\displaystyle \int_0^{\pi/2}\ln\sin\theta\;\mathrm d\theta=-\frac{\pi}{2}\ln2$

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$\ds{}$
\begin{align}&\color{#99f}{\large%
\int_{0}^{4}{\ln\pars{x} \over \root{4x – x^{2}}}\,\dd x}
=\int_{0}^{4}{\ln\pars{4\bracks{x/4}} \over \root{x/4 – \bracks{x/4}^{2}}}
\,{\dd x \over 4}
=\int_{0}^{1}{\ln\pars{4x} \over \root{x – x^{2}}}\,\dd x
\\[5mm]&=2\int_{0}^{1}{\pars{4x}^{-1/2}\,\ln\pars{4x}\pars{1- x}^{-1/2}}\,\dd x
=2\lim_{\mu\ \to\ -1/2}\,\,\,\partiald{}{\mu}
\int_{0}^{1}{\pars{4x}^{\mu}\pars{1- x}^{-1/2}}\,\dd x
\\[5mm]&=2\lim_{\mu\ \to\ -1/2}\,\,\,\partiald{}{\mu}\bracks{%
4^{\mu}\,{\Gamma\pars{\mu + 1}\Gamma\pars{1/2} \over \Gamma\pars{\mu + 3/2}}}
=\color{#66f}{\large 0}
\end{align}