Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How do I integrate this guy? I've been stuck on this for hours..

$$I=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}\frac{\ln(\sin y)\ln(\cos y)}{\sin y\cos y}dy$$

share|improve this question
7  
Is it really $dx$, and not $dy$? –  Brian M. Scott Dec 22 '12 at 4:35
4  
This earlier question is very relevant. –  Sasha Dec 22 '12 at 5:25
    
@Sasha, I enjoyed your answer. Though the calculation becomes slightly harder in this problem as the singularity arises for both $z$ and $w$ for $\frac{\partial^2}{\partial z \partial w}\beta(z, w)$, they all cancel out and yield the correct answer. –  sos440 Dec 22 '12 at 6:10
3  
The integral looks pretty sane to me. But really, we prefer to avoid subjective titles. –  Pedro Tamaroff Dec 22 '12 at 20:56

4 Answers 4

Let $\sin(y) = t$. Recall that $\cos^2(y) = 1-t^2$. Then $\cos(y) dy = dt$. Hence, $$I = \dfrac14 \int_0^1 \dfrac{\ln(\sin(y)) \ln(\cos^2(y)) \cos(y) dy}{\sin(y) \cos^2(y)}$$ $$4I = \int_0^1 \dfrac{\ln(t) \ln(1-t^2) dt}{t (1-t^2)}$$ $$\dfrac{\ln(t) \ln(1-t^2)}{t (1-t^2)} = -\dfrac{\ln(t)}{t(1-t^2)} \sum_{k=1}^{\infty} \dfrac{t^{2k}}k = - \sum_{k=1}^{\infty}\dfrac{\ln(t)}{(1-t^2)} \dfrac{t^{2k-1}}k$$ $$\dfrac{\ln(t)}{(1-t^2)} \dfrac{t^{2k-1}}k = \dfrac1k \sum_{l=0}^{\infty}t^{2k+2\ell-1} \ln(t)$$ Hence, $$\dfrac{\ln(t) \ln(1-t^2)}{t (1-t^2)} = -\sum_{k=1}^{\infty} \dfrac1k\sum_{l=0}^{\infty}t^{2k+2\ell-1} \ln(t)$$ Recall that $$\int_0^1 t^m \log(t) dt = -\dfrac1{(m+1)^2}$$ Hence, $$4I = \sum_{k=1}^{\infty} \sum_{\ell=0}^{\infty} \dfrac1k \dfrac1{(2k+2 \ell)^2} = \dfrac14 \sum_{k=1}^{\infty} \sum_{\ell=0}^{\infty} \dfrac1k \dfrac1{(k+\ell)^2} = \dfrac14 2 \zeta(3)$$ Hence, $$I = \dfrac{\zeta(3)}8$$

EDIT

Consider the sum $$S = \sum_{k=1}^{\infty} \sum_{l=0}^{\infty} \dfrac1k\dfrac1{(k+l)^2}$$ We have that $$S = \underbrace{\sum_{k=1}^{\infty} \sum_{l=k}^{\infty} \dfrac1k\dfrac1{l^2} = \sum_{l=1}^{\infty} \sum_{k=1}^l \dfrac1k\dfrac1{l^2}}_{\text{Change order of summation}} = \sum_{l=1}^{\infty} \dfrac{H_l}{l^2}$$ Also, note that $$S = \sum_{k=1}^{\infty} \dfrac1k\dfrac1{k^2} + \sum_{k=1}^{\infty} \sum_{l=1}^{\infty} \dfrac1k\dfrac1{(k+l)^2} = \zeta(3) + \sum_{k=1}^{\infty} \sum_{l=1}^{\infty} \dfrac1k\dfrac1{(k+l)^2}$$

Now let us evaluate the second sum in the above line.(To evaluate this, we follow the technique @sos440 has in his answer.)

\begin{align*} \sum_{k=1}^{\infty} \sum_{l=1}^{\infty} \frac{1}{k} \frac{1}{(k+l)^2} &= \frac{1}{2}\sum_{k,l=1}^{\infty} \left\{ \frac{1}{k} \frac{1}{(k+l)^2} + \frac{1}{l} \frac{1}{(k+l)^2} \right\} \\ &= \frac{1}{2}\sum_{k,l=1}^{\infty} \frac{1}{kl(k+l)} = \frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{k^2}\sum_{l=1}^{\infty} \left\{ \frac{1}{l} - \frac{1}{k+l}\right\} \\ &= \frac{1}{2}\sum_{k=1}^{\infty} \frac{H_{k}}{k^2} \end{align*}

Hence, we have $$S = \sum_{n=1}^{\infty} \frac{H_{n}}{n^2} = \zeta(3) + \frac{1}{2}\sum_{k=1}^{\infty} \frac{H_{k}}{k^2}$$ This gives us $$S = 2 \zeta(3)$$

share|improve this answer
4  
Why does the last double sum equal $2\zeta(3)$? –  Eric Naslund Dec 22 '12 at 7:03
2  
@EricNaslund Have added the reason. –  user17762 Dec 22 '12 at 20:47
1  
@EricNaslund Thanks for asking the question. Initially, I just evaluated it using wolfram-alpha, hoping that some trivial algebraic manipulation could give the answer. But it wasn't as trivial as I thought. Thanks. –  user17762 Dec 22 '12 at 20:54
    
@EricNaslund The fact that $\displaystyle \sum_{k=1}^{\infty} \dfrac{H_k}{k^2} = 2 \zeta(3)$ is an instance of the sum conjecture mentioned in the paper, equation $1$ and page $281$, being true for $n=3$ and $k=2$. Just wanted to share this nice information with you, in case you were not aware of it. –  user17762 Jan 13 '13 at 7:35

Let $x = \sin^2y$. (It seems we've started like @Sasha here and like @sos440.) Then $$\begin{eqnarray*} I &=& \frac{1}{16} \int_0^1 dx\, \frac{\log(x)\log(1-x)}{x(1-x)} \\ &=& \frac{1}{16} \int_0^1 dx\, \frac{\log(x)\log(1-x)}{x} + \underbrace{\frac{1}{16} \int_0^1 dx\, \frac{\log(x)\log(1-x)}{1-x}}_{x\to 1-x} \qquad (\textrm{partial fractions}) \\ &=& \frac{1}{8} \int_0^1 dx\, \frac{\log(x)\log(1-x)}{x} \qquad (\textrm{integral linked above}) \\ &=& \frac{1}{8} \int_0^1 dx\, \frac{\log(x)}{x} \left(-\sum_{k=1}^\infty \frac{x^k}{k}\right) \qquad (\textrm{Taylor expansion for }\log(1-x) ) \\ &=& -\frac{1}{8} \sum_{k=0}^\infty \frac{1}{k+1} \underbrace{\int_0^1 dx\, x^k \log x}_{-1/(k+1)^2} \qquad (\textrm{standard integral involving log}) \\ &=& \frac{1}{8} \sum_{k=0}^\infty \frac{1}{(k+1)^3} \\ &=& \frac{\zeta(3)}{8}. \end{eqnarray*}$$

Addendum: Note that $\csc(y)\sec(y) = \cot y + \tan y$. Then $$\begin{eqnarray*} I &=& \frac{1}{2} \int_{0}^{\pi/2} dy\, \frac{\ln(\sin y)\ln(\cos y)}{\sin y\cos y} \\ &=& \frac{1}{2} \int_{0}^{\pi/2} dy\, \frac{\ln(\sin y)\ln(\cos y)}{\tan y} + \underbrace{\frac{1}{2} \int_{0}^{\pi/2} dy\, \frac{\ln(\sin y)\ln(\cos y)}{\cot y}}_{y\to \pi/2-y} \\ &=& \int_{0}^{\pi/2} dy\, \frac{\ln(\sin y)\ln(\cos y)}{\tan y}. \end{eqnarray*}$$ This is exactly the integral linked above.

share|improve this answer
1  
I love this answer! (And of course +1 upvote.) My answer, compared to this concise and brilliant solution, seems just a dull sledgehammer. :( –  sos440 Dec 22 '12 at 8:55
    
Thanks @sos440. –  user26872 Dec 22 '12 at 18:31
1  
@oen, +1, very nice solution. –  Eric Naslund Dec 22 '12 at 20:49
1  
@oen +1. Neat.${}$ –  user17762 Dec 22 '12 at 20:51
    
@oen: by far this is the most brilliant answer! (+1) :D –  Chris's sis Dec 26 '12 at 9:43

I will assume that your $dx$ is indeed a type of $dy$, otherwise the answer is clear.

Let $t = \sin^2 y$. Then $dt = 2 \sin y \cos y \, dy$ and we have

$$I = \frac{1}{16} \int_{0}^{1} \frac{\log t \log(1-t)}{t(1-t)} \, dt. $$

Then thanks to the Taylor expansion

$$ \frac{\log(1-t)}{t} = -\sum_{n=1}^{\infty} \frac{t^{n-1}}{n}, $$

we have

\begin{align*} I &= -\frac{1}{16} \int_{0}^{1} \frac{\log t}{1-t} \sum_{n=1}^{\infty} \frac{t^{n-1}}{n} \, dt = -\frac{1}{16} \sum_{n=1}^{\infty} \frac{1}{n}\int_{0}^{1} \frac{t^{n-1} \log t}{1-t} \, dt \\ &= -\frac{1}{16} \sum_{n=1}^{\infty} \frac{1}{n} \sum_{m=0}^{\infty} \int_{0}^{1} t^{n+m-1} \log t \, dt \\ &= \frac{1}{16} \sum_{n=1}^{\infty} \sum_{m=0}^{\infty} \frac{1}{n} \frac{1}{(m+n)^2} \\ &= \frac{1}{16}\zeta(3) + \frac{1}{16}\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n} \frac{1}{(m+n)^2}. \end{align*}

Now we focus on the double summation in the last line. By interchanging the role of $m$ and $n$,

\begin{align*} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{1}{n} \frac{1}{(m+n)^2} &= \frac{1}{2}\sum_{m,n=1}^{\infty} \left\{ \frac{1}{n} \frac{1}{(m+n)^2} + \frac{1}{m} \frac{1}{(m+n)^2} \right\} \\ &= \frac{1}{2}\sum_{m,n=1}^{\infty} \frac{1}{mn(m+n)} = \frac{1}{2}\sum_{n=1}^{\infty} \frac{1}{n^2}\sum_{m=1}^{\infty} \left\{ \frac{1}{m} - \frac{1}{m+n}\right\} \\ &= \frac{1}{2}\sum_{n=1}^{\infty} \frac{H_{n}}{n^2} \end{align*}

To this end, it is natural to consider the following power series.

$$ \sum_{n=1}^{\infty} \frac{H_n}{n^2} x^n. $$

Associated with this series we easily observe that

$$ \sum_{n=1}^{\infty} H_n x^n = -\frac{\log(1-x)}{1-x}. $$

Thus we have

\begin{align*} \sum_{n=1}^{\infty} \frac{H_n}{n} x^n &= -\int_{0}^{x} \frac{\log(1-t)}{t(1-t)} \, dt = -\int_{0}^{x} \left\{ \frac{\log(1-t)}{t} + \frac{\log(1-t)}{1-t} \right\} \, dt \\ &= \mathrm{Li}_2 (x) + \frac{1}{2}\log^2(1-x), \end{align*}

where $\mathrm{Li}_s (x)$ is the polylogarithm defined by

$$ \mathrm{Li}_s (x) = \sum_{n=1}^{\infty} \frac{x^n}{n^s}. $$

Thus we must have

\begin{align*} \sum_{n=1}^{\infty} \frac{H_n}{n^2} x^n &= \int_{0}^{x} \frac{1}{t}\left\{ \mathrm{Li}_2 (t) + \frac{1}{2}\log^2(1-t) \right\} \, dt \\ &= \mathrm{Li}_3 (x) + \frac{1}{2} \int_{0}^{x} \frac{\log^2(1-t)}{t} \, dt. \end{align*}

Let $I(x)$ denote the latter integral. Then by integration by parts,

\begin{align*} 2I(x) = \int_{0}^{x} \frac{\log^2(1-t)}{t} \, dt = -\mathrm{Li}_2(x) \log(1-x) - \int_{0}^{x} \frac{\mathrm{Li}_2(t)}{1-t} \, dt. \end{align*}

But we know that $\mathrm{Li}_2 (x)$ satisfies the Euler reflection formula

$$ \mathrm{Li}_2 (x) + \mathrm{Li}_2 (1-x) = \zeta(2) - \log x \log (1-x), $$

which is easily proved by differentiation both sides with respect to $x$. Thus we have

\begin{align*} \int_{0}^{x} \frac{\mathrm{Li}_2 (t)}{1-t} \; dt & = \int_{0}^{x} \frac{1}{1-t} \left[ \zeta(2) - \log t \log (1-t) - \mathrm{Li}_2 (1-t) \right] \; dt \\ & = \int_{0}^{x} \frac{\zeta(2) - \mathrm{Li}_2 (1-t)}{1-t} \; dt - \int_{0}^{x} \frac{\log t \log (1-t)}{1-t} \; dt. \end{align*}

But the fist integral is easily calculated as we can find its antiderivative:

\begin{align*} \int_{0}^{x} \frac{\zeta(2) - \mathrm{Li}_2 (1-t)}{1-t} \; dt &= \left[ -\zeta(2)\log(1-t) + \mathrm{Li}_{3} (1-t) \right]_{0}^{x} \\ &= -\zeta(2) \log(1-x) + \mathrm{Li}_{3}(1-x) - \zeta(3). \end{align*}

The second integral reduces to a formula involving $I(x)$:

\begin{align*} -\int_{0}^{x} \frac{\log t \log (1-t)}{1-t} \; dt &= \left[ \frac{1}{2} \log^2 (1-t) \log t \right]_{0}^{x} - I(x) \\ &= \frac{1}{2} \log^2 (1-x) \log x - I(x). \end{align*}

From this calculation, we obtain a closed form for $I(x)$ as follows:

$$ I(x) = \zeta(3) - \mathrm{Li}_3 (1-x) + \{ \zeta(2) - \mathrm{Li}_2 (x) \} \log (1 - x) - \frac{1}{2} \log^2 (1-x) \log x. $$

Plugging $x = 1$, we then have

$$ I(1) = \zeta(3).$$

Therefore we have

$$ \sum_{n=1}^{\infty} \frac{H_n}{n^2} = 2 \zeta(3)$$

and hence

\begin{align*} I = \frac{1}{16}\zeta(3) + \frac{1}{16} \zeta(3) = \frac{1}{8}\zeta(3). \end{align*}

share|improve this answer
1  
+1. Though evaluation of $$\sum_{k=1}^{\infty} \sum_{l=0}^{\infty} \dfrac1k \dfrac1{(k+l)^2}$$ could be simplified a lot. See my post. –  user17762 Dec 22 '12 at 20:48

According to Mathematica, $$I=\frac{\zeta(3)}{8}$$where $\zeta$ is the zeta function. (See also this about the integral). This is approximately 0.150257.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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