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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ \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

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 $\;\displaystyle \int_0^{\pi/2}\ln\sin\theta\;\mathrm d\theta=-\frac{\pi}{2}\ln2$

  • 1
    $\begingroup$ @MathGod Thank you ^^ $\endgroup$ – Venus Dec 23 '14 at 7:36

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




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




Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.