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Q: Evaluate$\newcommand{\dx}{\mathrm dx}\newcommand{\du}{\mathrm du}\newcommand{\dv}{\mathrm dv}\newcommand{\dtheta}{\mathrm d\theta}\newcommand{\dw}{\mathrm dw}$$$I=\int\limits_0^{\pi/2}\dx\,\frac {\left(\log\sin x\log\cos x\right)^2}{\sin x\cos x}$$

I'm out of ideas on what to do. I tried using the astute limit identity for integration, but that lead to nowhere, because it's simply the same integral i.e adding the two together, doesn't yield a simplification whatsoever$$I=\int\limits_0^{\pi/2}\dx\,\frac {\log^2\sin x\log^2\cos x}{\sin x\cos x}=\int\limits_0^{\pi/2}\du\,\frac {\log^2\cos u\log^2\sin u}{\cos u\sin u}$$I've looked into making a u-substitution, but have no idea where to begin. I plan to use a trigonometric identity, namely$$\sec x\csc x=\cot x+\tan x$$but I'm not sure what to do afterwards because we get$$I=\int\limits_0^{\pi/2}\dx\,\cot x\log^2\sin x\log^2\cos x+\int\limits_0^{\pi/2}\dx\,\tan x\log^2\sin x\log^2\cos x$$I would appreciate it if you guys gave me an idea on where to begin!

Also, as a side note, I've used \newcommand on \dx,\du,\dv, and \dw to automatically change to $\dx$, $\du$, $\dv$, and $\dw$ respectively, if you guys don't mind! Just to make it easier for the differential operator!

Crescendo
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2 Answers2

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Change variable to $t = \sin^2 x$ and notice $$\begin{align}\frac{dx}{\sin x\cos x} &= \frac{\sin x\cos x dx}{\sin^2 x\cos^2 x} = \frac12\frac{dt}{t(1-t)}\\ \log\sin x \log\cos x &= \frac14\log\sin^2 x \log \cos^2 x = \frac14 \log t\log(1-t) \end{align} $$ The integral at hand can be rewritten as $$\mathcal{I} \stackrel{def}{=} \int_0^{\pi/2} \frac{(\log\sin x\log\cos x)^2}{\sin x\cos x} dx = \frac{1}{32}\int_0^1 \frac{\log^2 t \log^2(1-t)}{t(1-t)} dt$$ Since $\displaystyle\frac{1}{t(1-t)} = \frac1t + \frac{1}{1-t}$, by replacing $t$ by $1-t$ in part of the integral, we find

$$\begin{align} \mathcal{I} &= \frac{1}{16}\int_0^1 \frac{\log^2 t\log^2(1-t)}{t} dt = \frac{1}{48}\int_0^1 \log^2(1-t) d\log^3 t\\ &\stackrel{\text{I.by.P}}{=} \frac{1}{48}\left\{ \left[\log^2(1-t)\log^3 t\right]_0^1 + 2 \int_0^1 \log^3 t\frac{\log(1-t)}{1-t}dt\right\}\\ &= -\frac{1}{24}\int_0^1 \frac{\log^3 t}{1-t}\sum_{n=1}^\infty\frac{t^n}{n} dt = -\frac{1}{24}\int_0^1 \log^3 t\sum_{n=1}^\infty H_nt^n dt\\ &\stackrel{t = e^{-y}}{=} \frac{1}{24} \int_0^\infty y^3 \sum_{n=1}^\infty H_n e^{-(n+1)y} dy = \frac14 \sum_{n=1}^\infty \frac{H_n}{(n+1)^4} \end{align} $$ where $H_n$ are the $n^{th}$ harmonic number. By rearranging its terms, the last sum should be expressible in terms of zeta functions. I'm lazy, I just ask WA to evaluate the sum. As expected, last sum equals to $2\zeta(5) - \frac{\pi^2}{6}\zeta(3)$${}^\color{blue}{[1]}$.

As a consequence, the integral at hand equals to:

$$\mathcal{I} = \frac{12\zeta(5) - \pi^2\zeta(3)}{24} \approx 0.024137789997360933616411382857235691008...$$

Notes

  • $\color{blue}{[1]}$ It turns out we can compute this sum using an identity by Euler. $$2\sum_{n=1}^\infty \frac{H_n}{n^m} = (m+2)\zeta(m+1) - \sum_{n=1}^{m-2}\zeta(m-n)\zeta(n+1),\quad\text{ for } m = 2, 3, \ldots$$ In particular, for $m = 4$, this identity becomes $$\sum_{n=1}^\infty \frac{\zeta(n)}{n^4} = 3\zeta(5) - \zeta(2)\zeta(3)$$ and we can evaluate our sum as $$\begin{align}\mathcal{I} &= \frac14\sum_{n=1}^{\infty}\frac{H_{n}}{(n+1)^4} = \frac14\sum_{n=1}^{\infty}\left(\frac{H_{n+1}}{(n+1)^4}-\frac{1}{(n+1)^5}\right) = \frac14\left(\sum_{n=1}^\infty \frac{H_n}{n^4} - \zeta(5)\right)\\ &= \frac14(2\zeta(5) - \zeta(2)\zeta(3)) = \frac{12\zeta(5) - \pi^2\zeta(3)}{24} \end{align} $$
achille hui
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  • How did you get to the step below?$$\mathcal{I} = \frac{1}{16}\int_0^1 \frac{\log^2 t\log^2(1-t)}{t} dt = \frac{1}{48}\int_0^1 \log^2(1-t) d\log^3 t$$ – Crescendo Jan 17 '18 at 20:03
  • @Crescendo $$\begin{align}&\int_0^1\frac{\log^2(t)\log^2(1-t)}{t(1-t)}dt = \int_0^1\log^2(t)\log^2(1-t)\left(\frac{1}{t} + \frac{1}{1-t}\right)dt\ \stackrel{s=1-t}{=} & \int_0^1\frac{\log^2(t)\log^2(1-t)}{t}dt + \int_0^1\frac{\log^2(1-s)\log^2(s)}{s}ds = 2\int_0^1\frac{\log^2(t)\log^2(1-t)}{t}dt \end{align} $$ and $$d\log^3(t) = \frac{d(\log^3(t))}{dt} dt = 3\log^2(t)\frac{d\log(t)}{dt} dt = 3 \frac{\log^2(t)}{t}dt$$ – achille hui Jan 17 '18 at 23:23
  • So you're using differentiation under the integral? I'm having trouble understanding your notation $d,\log^3t$ for some reason... – Crescendo Jan 17 '18 at 23:53
  • @Crescendo For your purpose, you can either view $\int f(x)dg(x)$ as a short-hand for $\int f(x)g'(x)dx$ or as a Riemann–Stieltjes integral. I personally view those things as integration involving differential forms. (which have richer algebraic properties and allow more flexiable manipulation of the expression) – achille hui Jan 18 '18 at 00:54
  • One last question: How does$$\frac {1}{1-t}\sum\limits_{n\geq1}\frac {t^n}n=\sum\limits_{n\geq1}H_n t^n$$I think you represented the $1/(1-t)$ fraction as its infinite geometric sequence, but I'm not seeing how that equals the right-hand side. – Crescendo Feb 10 '18 at 02:03
  • @Crescendo In general, if function $f(t)$ has a power series representation $f(t) = \sum_{n=0}^\infty a_n t^n$ over some neighbor of $t = 0$, then $\frac{f(t)}{1-t}$ has power series representation $\sum_{n=0}^\infty (\sum_{k=0}^n a_k) t^n$. Above identity is a special case of that. – achille hui Feb 10 '18 at 05:29
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I will work with the integral found after a substitution of $t = \sin^2 x$ has been made so that $$I = \int_0^{\pi/2} \frac{\ln^2 (\sin x) \ln^2 (\cos x)}{\sin x \cos x} \, dx = \frac{1}{32} \int_0^1 \frac{\ln^2 t \ln^2 (1 - t)}{t (1 - t)} \, dt.$$

The first method that comes to mind is to evaluate the integral appearing to the right as the derivative of a beta function. Since $$\text{B} (x,y) = \int_0^1 t^{x - 1} (1 - t)^{y - 1} \, dt,$$ we see that $$\lim_{x \to 0^+} \lim_{y \to 0^+} \partial^2_x \partial^2_y \text{B}(x,y) = \int_0^1 \frac{\ln^2 t \ln^2 (1 - t)}{t (1 - t)} \, dt.$$ Thus $$I = \frac{1}{32} \lim_{x \to 0^+} \lim_{y \to 0^+} \partial^2_x \partial^2_y \text{B}(x,y).$$ While the above calculation is doable, it is far from trivial (at least "by hand") and will accordingly not be given here.

As a second method, I will make use of the following Maclaurin series expansion for the term $\ln^2 (1 - t)$ which is given by $$\ln^2 (1 - t) = 2 \sum_{n = 2}^\infty \frac{H_{n - 1}}{n} t^n, \qquad |t| < 1, \tag1$$ where $H_n$ is the $n$th Harmonic number. A proof of this result can be found here.

As $$\frac{1}{t(1 - t)} = \frac{1}{t} + \frac{1}{1 - t},$$ we can write the integral $I$ as $$I = \frac{1}{32} \int_0^1 \frac{\ln^2 t \ln^2 (1 - t)}{t} \, dt + \frac{1}{32} \frac{\ln^2 t \ln^2 (1 - t)}{1 - t} \, dt.$$ If in the second of these integrals we enforce a substitution of $t \mapsto 1 - t$ we have $$I = \frac{1}{16} \int_0^1 \frac{\ln^2 t \ln^2 (1 - t)}{t} \, dt.$$

Replacing the term $\ln^2(1 - t)$ with its Maclaurin series expansion, after interchanging the summation with the integral sign we have $$I = \frac{1}{8} \sum_{n = 2}^\infty \frac{H_{n - 1}}{n} \int_0^1 t^{n - 1} \ln^2 t \, dt.$$ The integral that results can be readily found using integration by parts twice. The result is $$\int_0^1 t^{n - 1} \ln^2 t \, dt = \frac{2}{n^3}.$$ Thus $$I = \frac{1}{4} \sum_{n = 2}^\infty \frac{H_{n - 1}}{n^4}.$$

Since $$H_n = H_{n - 1} + \frac{1}{n},$$ we have $$I = \frac{1}{4} \left [\sum_{n = 2}^\infty \frac{H_n}{n^4} - \sum_{n = 2}^\infty \frac{1}{n^5} \right ] = \frac{1}{4} \left [\sum_{n = 1}^\infty \frac{H_n}{n^4} - \sum_{n = 1}^\infty \frac{1}{n^5} \right ].$$

The value for the Euler sum is well-known (see here for example). It is $$\sum_{n = 1}^\infty \frac{H_n}{n^4} = 3 \zeta (5) - \zeta (2) \zeta (3),$$ while the second sum is just the zeta function $\zeta (5)$. Thus $$I = \frac{1}{4} \big{[}3 \zeta (5) - \zeta (2) \zeta (3) - \zeta (5) \big{]},$$ or $$\int_0^{\pi/2} \frac{\ln^2 (\sin x) \ln^2 (\cos x)}{\sin x \cos x} \, dx = \frac{1}{4} \big{[} 2 \zeta (5) - \zeta (2) \zeta (3) \Big{]}.$$

omegadot
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