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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.

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  • $\begingroup$ Wolfram's integrator gives a non elementary primitive. $\endgroup$ – ajotatxe Dec 13 '14 at 8:49
  • $\begingroup$ @ajotatxe The answer to this is 0 , but I don't know how to get it $\endgroup$ – M.S.E Dec 13 '14 at 8:50
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    $\begingroup$ \begin{align} \int^4_0\frac{\ln{x}}{\sqrt{x(4-x)}}{\rm d}x =&\int^1_0\frac{2\ln{2}+\ln{x}}{\sqrt{x(1-x)}}{\rm d}x\\ =&2\pi\ln{2}+\pi(-\gamma-2\ln{2}+\gamma)\\ =&0 \end{align} $\endgroup$ – M.N.C.E. Dec 13 '14 at 8:59
  • $\begingroup$ Try Subsituting $$u=\frac{x}{4}$$ $\endgroup$ – The Artist Dec 13 '14 at 9:02
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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, $\pi \log 2 -\pi\log 2$.

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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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    $\begingroup$ @MathGod Thank you ^^ $\endgroup$ – Venus Dec 23 '14 at 7:36
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{}$ \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}

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0
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$\int_{0}^{4}\frac{\ln{x}}{\sqrt{4x-x^2}}dx$

$=\int_{-2}^{2}\frac{\ln{(t+2)}}{\sqrt{4-t^2}}dt$

$=\frac12\int_{-2}^{2}\frac{\ln{(t+2)}}{\sqrt{4-t^2}}dt +\frac12\int_{-2}^{2}\frac{\ln{(-t+2)}}{\sqrt{4-t^2}}dt$

$=\frac12\int_{-2}^{2}\frac{\ln{(4-t^2)}}{\sqrt{4-t^2}}dt$

$=0$

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