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Finding $\displaystyle \int^{\frac{\pi}{2}}_{0}\ln(\sin x)\cdot \sin xdx$

What I try:-> Integration by parts

assuming $\displaystyle I = \int\ln(\sin x)\cdot \sin xdx = -\ln(\sin x)\cdot \cos x+\int\frac{\cos^2 x}{\sin x}dx$

$\displaystyle I = -\ln(\sin x)\cdot \cos x+\int\frac{1-\sin^2 x}{\sin x}dx$

$ = -\ln(\sin x)\cos x+\ln\bigg(\tan\frac{x}{2}\bigg)-\cos x$

$ \displaystyle \int^{\frac{\pi}{2}}_{0}\ln(\sin x)\cos xdx = \bigg[-\ln(\sin x)\cos x+\ln\bigg(\tan\frac{x}{2}\bigg)-\cos x\bigg]\bigg|^{\frac{\pi}{2}}_{0}=-\ln(0)+\ln(0)$

but answer is $\ln(2/e)$

could some explain me why I have got wrong answer,thanks

also explain me How I solve it using double integral

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    $\begingroup$ $\log(0)$ is not a number. $\endgroup$ Oct 15, 2018 at 15:00
  • $\begingroup$ Thanks Jack D'Aurizio, we have to write it as $\lim_{x\rightarrow 0}\ln(\sin x)\cdot \cos x$ $\endgroup$
    – DXT
    Oct 15, 2018 at 15:01
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    $\begingroup$ It still does not exist. $\lim(a-b) = \lim a-\lim b$ only if both $\lim a$ and $\lim b$ make sense. $\endgroup$ Oct 15, 2018 at 15:03

4 Answers 4

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\begin{align} I&=\int^{\frac{\pi}{2}}_{0}\ln(\sin x)\cdot \sin x\,dx \tag{1}\label{1} \end{align}

\begin{align} I&=\int^{\frac{\pi}{2}}_{0}\tfrac12\ln(\sin^2 x)\cdot \sin x\,dx \tag{2}\label{2} \\ &= \int^{\frac{\pi}{2}}_{0} \tfrac12\ln(1-\cos^2 x)\cdot \sin x\,dx \tag{3}\label{3} . \end{align}

Let $t=\cos x$, then we have

\begin{align} I&=\tfrac12\int_0^1\ln(1-t^2)\,dt \\ &= \tfrac12\int_0^1\ln(1-t)+\ln(1+t)\,dt \\ &= \left.\tfrac12 ( 1-t-(1-t)\ln(1-t) +(t+1)\ln(t+1)-1-t )\right|_0^1 =\ln2-1 . \end{align}

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  • $\begingroup$ This is the calculus I answer that the user should've gotten in the first place xD. $\endgroup$
    – user459879
    Oct 15, 2018 at 15:32
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$\log\sin x$ has a well-known Fourier series: $$ \log\sin x=-\log 2-\sum_{k\geq 1}\frac{\cos(2k x)}{k} $$ and for any $k\in\mathbb{N}^+$ we have $$ \int_{0}^{\pi/2}\cos(2kx)\sin(x)\,dx = -\frac{1}{(2k-1)(2k+1)}, $$ hence $$ \int_{0}^{\pi/2}\sin(x)\log\sin(x)\,dx = -\log(2)+\sum_{k\geq 1}\frac{1}{(2k-1)k(2k+1)} $$ where the last series equals $-1+2\log 2$ by partial fraction decomposition. It follows that $$ \int_{0}^{\pi/2}\sin(x)\log\sin(x)\,dx = \log(2)-1 $$ as wanted.

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  • $\begingroup$ Thanks Jack D'Aurizio. caoul you please explain me how i find $\displaystyle \int^{\frac{\pi}{2}}_{0}\cos(2kx)\cdot \sin (x)dx =\frac{1}{(1+2k)(1-2k)}.$ $\endgroup$
    – DXT
    Oct 15, 2018 at 15:03
  • $\begingroup$ @DurgeshTiwari: Sine addition formulas and explicit integration. $\endgroup$ Oct 15, 2018 at 15:04
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    $\begingroup$ It's not that I don't like the elegance of your answer, it's just that the question is probably from a Calc II or Calc III class where, much of the time, Fourier series haven't been properly introduced. $\endgroup$
    – Sam
    Oct 15, 2018 at 15:05
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    $\begingroup$ @Leo: that is not my fault. According to my opinion, Fourier series should be introduced as soon as possible, since they provide multiple ways for explicit evaluations, like in this case or in Basel problem. I also do not believe this exercise comes from a Calc-X class: the OP is asking for elementary integrals from quite some time. $\endgroup$ Oct 15, 2018 at 15:07
  • $\begingroup$ Why is this called a 'Fourier' series? Is that just the name for any trigonometric series? or can one find it using the classic method? $\endgroup$
    – clathratus
    Jan 12, 2019 at 23:34
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Other answers are good but I prefer to talk about yours. You found (with a typo) \begin{align} \int_{0}^{\frac{\pi}{2}}\ln(\sin x)\ \sin x\ dx &= -\ln(\sin x)\cos x+\ln\bigg(\tan\frac{x}{2}\bigg)\color{red}{+}\cos x\Big|_{0}^{\frac{\pi}{2}} \\ &= 0 + \lim_{x\to0}\bigg(\ln(\sin x)\cos x+\ln\tan\frac{x}{2}\bigg)-1 \\ &= 0 + \lim_{x\to0}\bigg(\ln(1+\cos x)-(1-\cos x)\ln\sin x\bigg)-1 \\ &= \ln2-1 \end{align}

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Here is an approach following along lines similar to your own answer. There is however a small subtlety used in the first integration by parts step.

On integrating by parts, we have $$\int_0^{\frac{\pi}{2}} \sin x \ln (\sin x) \, dx = (1 - \cos x) \ln (\sin x) \Big{|}_0^{\pi/2} - \int_0^{\frac{\pi}{2}} (1 - \cos x) \cdot \frac{\cos x}{\sin x} \, dx.$$ Note the subtlety here. Having chosen $v' = \sin x$ we have used $v = 1 - \cos x$, that is, a non-zero constant of integration has been selected. Doing so means one has zero at the upper and lower limits of integration.

Continuing, we have \begin{align} \int_0^{\frac{\pi}{2}} \sin x \ln (\sin x) \, dx &= \int_0^{\frac{\pi}{2}} \frac{-\cos x + \cos^2 x}{\sin x} \, dx\\ &= \int_0^{\frac{\pi}{2}} \frac{-\cos x + 1 - \sin^2 x}{\sin x} \, dx\\ &= \int_0^{\frac{\pi}{2}} \left [\text{cosec} \, x - \cot x - \sin x \right ] \, dx\\ &= \left [-\ln (\text{cosec} \,x + \cot x) - \ln (\sin x) + \cos x \right ]_0^{\pi/2}\\ &= \left [-\ln (1 + \cos x) + \cos x \right ]_0^{\pi/2}\\ &= \ln 2 - 1, \end{align} as expected.

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