4
$\begingroup$

Problem: Evaluate:

$$I=\int_0^{\pi/2} \ln(\sin(x))\tan(x)dx$$

I tried to attempt it by using the Beta, Gamma and Digamma Functions. My approach was as follows:

$$$$ Consider $$I(a,b)=\int_0^{\pi/2} \sin^a(x)\sin^b(x)\cos^{-b}(x)dx$$ $$$$ $$=\dfrac{1}{2}\beta\bigg (\dfrac{a+b+1}{2},\dfrac{1-b}{2}\bigg )= \dfrac{1}{2} \dfrac{\Gamma(\frac{a+b+1}{2})\Gamma(\frac{1-b}{2})}{\Gamma(\frac{a+2}{2})}$$ Now, $$\dfrac{1}{2}\dfrac{\partial}{\partial a} \beta\bigg (\dfrac{a+b+1}{2},\dfrac{1-b}{2}\bigg )\bigg |_{a=0,b=1} = \int_0^{\pi/2}\ln(\sin(x))\sin^a(x)\tan^b(x)dx \bigg |_{a=0,b=1} = I$$ Now, $$\dfrac{\partial}{\partial a} \beta\bigg (\dfrac{a+b+1}{2},\dfrac{1-b}{2}\bigg )$$ $$$$ $$=\dfrac{1}{2} \dfrac{\Gamma(\frac{1-b}{2})}{(\Gamma(\frac{a+2}{2}))^2}\bigg (\Gamma '\bigg (\frac{a+b+1}{2}\bigg )\Gamma \bigg ( \frac{a+2}{2}\bigg ) - \Gamma '\bigg ( \frac{a+2}{2}\bigg )\Gamma \bigg (\frac{a+b+1}{2}\bigg )\bigg ) $$ $$$$ $$= \dfrac{1}{2}\dfrac{\Gamma(\frac{1-b}{2})}{\Gamma(\frac{a+2}{2})}\bigg (\psi\bigg ( \frac{a+b+1}{2}\bigg )\Gamma \bigg (\frac{a+b+1}{2}\bigg ) - \psi\bigg ( \frac{a+2}{2}\bigg )\Gamma \bigg (\frac{a+b+1}{2}\bigg )\bigg )$$ $$$$ $$=\dfrac{1}{2} \dfrac{\Gamma(\frac{a+b+1}{2})\Gamma(\frac{1-b}{2})}{\Gamma(\frac{a+2}{2})}\bigg (\psi\bigg ( \frac{a+b+1}{2}\bigg )-\psi\bigg ( \frac{a+2}{2}\bigg )\bigg )$$ $$$$ $$\Rightarrow\dfrac{\partial}{\partial a} \beta\bigg (\dfrac{a+b+1}{2},\dfrac{1-b}{2}\bigg )=\dfrac{1}{2} \dfrac{\Gamma(\frac{a+b+1}{2})\Gamma(\frac{1-b}{2})}{\Gamma(\frac{a+2}{2})}\bigg (\psi\bigg ( \frac{a+b+1}{2}\bigg )-\psi\bigg ( \frac{a+2}{2}\bigg )\bigg )$$

$$$$

$$\Longrightarrow \dfrac{1}{2} \dfrac{\partial}{\partial a} \beta \bigg ( \dfrac{a+b+1}{2} ,\dfrac{1-b}{2} \bigg ) \bigg |_{a=0,b=1} = I $$ $$$$ $$=\dfrac{1}{2}\times \dfrac{1}{2} \dfrac{\Gamma(\frac{a+b+1}{2})\Gamma(\frac{1-b}{2})}{\Gamma(\frac{a+2}{2})}\bigg (\psi\bigg ( \frac{a+b+1}{2}\bigg )-\psi\bigg ( \frac{a+2}{2}\bigg )\bigg ) \bigg |_{a=0,b=1}$$ $$$$ $$ =\dfrac{1}{4}\dfrac{\Gamma(0)\Gamma(1)}{\Gamma(1)}\bigg (\psi\bigg ( 1 \bigg )-\psi\bigg (1\bigg )\bigg )$$ $$$$ Could somebody please be so kind as to tell me where I have gone wrong? I would be truly grateful for your assistance. Thanks very, very much in advance!

$\endgroup$
  • $\begingroup$ Not an answer, but isn't it easier to start with $u=\sin t$? $\endgroup$ – mickep Jun 14 '15 at 8:17
  • $\begingroup$ Probably Sir. but actually I tend to try Differentiating Under the Integral Sign whenever I see $\ln$ integrals. It's just since I'm far more comfortable with Differentiation Under the Integral Sign as compared to getting the right substitution.$$$$ EDIT: Another factor is that I recently discovered for myself that suppose we have an integrand containing $\ln$, then we can consider any function such that when we take its partial derivatives and plug in certain values, then we get the Integrand. I was really excited by this (small) 'discovery' and have been trying it in every question. $\endgroup$ – Ishan Jun 14 '15 at 8:20
  • 1
    $\begingroup$ Your calculations are correct. Evaluate the derivative at $a=0$ and take the limit when $b\rightarrow 1$ $\endgroup$ – d.k.o. Jun 14 '15 at 9:12
  • $\begingroup$ Sir, please could you show me how I am to take the limit? I was unable to understand. $$$$PS. I was thrilled to read that they were indeed correct! $\endgroup$ – Ishan Jun 14 '15 at 9:12
  • $\begingroup$ $\dfrac{1}{4}\dfrac{\Gamma(\frac{a+b+1}{2})\Gamma(\frac{1-b}{2})}{\Gamma(\frac{a+2}{2})}\bigg (\psi\bigg ( \frac{a+b+1}{2}\bigg )-\psi\bigg ( \frac{a+2}{2}\bigg )\bigg ) \bigg |_{a=0}= \dfrac{\pi}{4}\Big(\Psi\Big(\frac{1+b}{2}\Big)+\gamma\Big)\times \sin^{-1}\Big(\pi\Big(\frac{1+b}{2}\Big)\Big)$ $\endgroup$ – d.k.o. Jun 14 '15 at 9:20
4
$\begingroup$

OK, I realize this does not really answer your question (but look at the update below), in the sense it does not point at your error. Anyways, I think that you should be flexible with your methods doing integrals, so here is how it can be done with the change of variables I suggested:

You end up with the integral $$ I=\int_0^1 \log u\frac{u}{1-u^2}\,du. $$ Now writing $$ \frac{1}{1-u^2}=1+u^2+u^4+\cdots, $$ and using that (integrating by parts) $$ \begin{aligned} \int_0^1 u^{2k+1}\log u\,du&=\bigl[\frac{u^{2k+2}}{2k+2}\log u\bigr]_0^1-\frac{1}{2k+2}\int_0^1 u^{2k+2}\frac{1}{u}\,du\\ &=-\frac{1}{4}\frac{1}{(k+1)^2}. \end{aligned} $$ Thus, $$ I=-\frac{1}{4}\sum_{k=0}^{+\infty}\frac{1}{(k+1)^2}=-\frac{\pi^2}{24}. $$ The last equality by the Basel problem.

Updated version

The problem in your calculation, by inserting $a=0$ and $b=1$, you get a $\Gamma(0)$ (which is undefined, or as its best $\pm\infty$ depending on if you let $b\to 1$ from left or right) times something that is zero. This is indeterminate, and you must look at limits. One way is as follows:

If you first let $a=0$, and write $b=1+\epsilon$, then you get $$ \frac{1}{4}\Gamma(-\epsilon/2)\Gamma(1+\epsilon/2)\bigl(\Psi(1+\epsilon/2)-\Psi(1)\bigr) $$ We multiply and divide by $\epsilon/2$ (lethal weapon number 2), to write this as $$ \frac{1}{4}\frac{\epsilon}{2}\Gamma(-\epsilon/2)\Gamma(1+\epsilon/2)\frac{\Psi(1+\epsilon/2)-\Psi(1)}{\epsilon/2} $$ By letting $\epsilon\to 0$, you will find that the limit is $$ \frac{1}{4}\times (-1)\times 1 \times \frac{\pi^2}{6}=-\frac{\pi^2}{24}. $$ Here, we have used the facts that $\lim_{x\to 0} x\Gamma(x)=1$ and $\Psi^{(1)}(1)=\pi^2/6$.

$\endgroup$
  • $\begingroup$ Thanks you Sir. I completely agree with your point on being flexible. However, I faced a similar problem even when trying to integrate $$\int_0^{\pi/2}\ln(\sin(x))\ln(\tan(x))dx$$Unless I rectify this error , I fear I'm making a significant conceptual error. Sir, please could you take the time to check my original problem? I tried for the past 30 minutes, but still couldn't spot the error. $\endgroup$ – Ishan Jun 14 '15 at 9:09
  • $\begingroup$ I'm cooking food with a baby on my stomach, so I won't have time right now... Hopefully this afternoon. If no one else does it before, of course. $\endgroup$ – mickep Jun 14 '15 at 9:12
  • $\begingroup$ Anytime that suits you, Sir. Many many thanks in advance! Enjoy the meal you'll have:) $\endgroup$ – Ishan Jun 14 '15 at 9:13
  • $\begingroup$ I have updated the answer with details for the real question. Also, the integral that you mention with two logarithms looks more interesting. My CAS tells me a value, but I don't see how to get it at the moment... $\endgroup$ – mickep Jun 14 '15 at 10:58
  • $\begingroup$ Alright Sir, thanks very very much! $$$$ Sir, I reached a very similar stage with that Integral too using almost the same method. The only change was that I had to take $$\dfrac{\partial^2}{\partial a\partial b} \dfrac{1}{2} \beta \bigg (\dfrac{a+b+1}{2},\dfrac{1-b}{2} \bigg )\bigg |_{a=b=0}$$ $\endgroup$ – Ishan Jun 14 '15 at 11:35
0
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} I & = \int_{0}^{\pi/2}\ln\pars{\sin\pars{x}}\tan\pars{x}\,\dd x \,\,\,\stackrel{x\ =\ \arcsin\pars{t}}{=}\,\,\, \int_{0}^{1}\ln\pars{t}{t \over \root{1 - t^{2}}} \,{\dd t \over \root{1 - t^{2}}} \\[5mm] & = \int_{0}^{1}\ln\pars{t}\,{t \over 1 - t^{2}}\,\dd t = \int_{0}^{1}\ln\pars{t}\pars{{1/2 \over 1 - t} - {1/2 \over 1 + t}}\,\dd t \\[5mm] & = -\,{1 \over 2}\int_{0}^{1}\bracks{-\,{{\ln\pars{1 - t} \over t}}}\,\dd t + {1 \over 2}\int_{0}^{-1}{\ln\pars{-t} \over 1 - t}\,\dd t \\[5mm] & = -\,{1 \over 2}\int_{0}^{1}\bracks{-\,{{\ln\pars{1 - t} \over t}}}\,\dd t - {1 \over 2}\int_{0}^{-1}\bracks{-\,{\ln\pars{1 - t} \over t}}\,\dd t \\[5mm] & = -\,{1 \over 2}\int_{0}^{1}\mrm{Li}_{2}'\pars{t}\,\dd t - {1 \over 2}\int_{0}^{-1}\mrm{Li}_{2}'\pars{t}\,\dd t = -\,{1 \over 2}\bracks{\mrm{Li}_{2}\pars{1} + \mrm{Li}_{2}\pars{-1}} \\[5mm] & = -\,{1 \over 2}\sum_{n = 1}^{\infty} \bracks{{1 \over n^{2}} + {\pars{-1}^{n} \over n^{2}}} = -\,{1 \over 2}\sum_{n = 1}^{\infty} \bracks{{1 \over \pars{2n}^{2}} + {1 \over \pars{2n}^{2}}} = -\,{1 \over 4}\ \underbrace{\sum_{n = 1}^{\infty}{1 \over n^{2}}} _{\ds{\pi^{2} \over 6}} = \bbx{-\,{\pi^{2} \over 24}} \end{align}

$\endgroup$

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.