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Hi I am trying to integrate a log trigonometric integral given by $$ I:=\int_0^{\pi/2}\log^4 \tan \frac{x}{2}dx=\frac{5\pi^5}{32}. $$ This is very similar to a previous integral posted except the power of the logarithm. Note this integral is also equal to $$ \int_0^{\pi/2}\log^4 \tan x \, dx=\frac{5\pi^5}{32}. $$ I have wrote I as $$ I=\int_0^{\pi/2} \left(\log \sin \frac{x}{2}-\log \cos \frac{x}{2}\right)^4 dx $$ but got stuck here since factoring this out seems like a mess. Having seen how David H solved a similar integral I posted, I tried another method starting with I and using $t=\tan x/2$, and obtained $$ I=2\int_0^{1}\log^4 t \frac{dt}{1+t^2}. $$ Following this I tried $u=-\log t$ but got stuck after this. Thanks, it would be nice to see a solution that doesn't reduce the integral to a difficult sum to evaluate

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    $\begingroup$ It is similar to this one. $\endgroup$ – Felix Marin Apr 27 '14 at 4:17
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    $\begingroup$ Try expanding the $1/(1+t^2)$ using a geometric series, then using the result that $$ \int_0^1 \log^4(t) t^{2n} \ \mathrm{d}t = \frac{4!}{(2n+1)^5}. $$ Edit - just saw the caveat about difficult sums. Nevermind. $\endgroup$ – Bennett Gardiner Apr 27 '14 at 4:24
  • $\begingroup$ @BennettGardiner Yes, thank you for the suggestion though! It is nice to see it done another way than having to evaluate a non-trivial sum at the end. $\endgroup$ – Jeff Faraci Apr 27 '14 at 4:56
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ By following my previous answer you'll arrive to \begin{align} &\half\,\lim_{\mu \to 0}\partiald[4]{\sec\pars{\pi\mu/2}}{\mu}={\pi \over 2}\times \\[3mm]&\lim_{\mu \to 0}\bracks{% {5 \over 16}\,\pi^{4}\sec^{5}\left(\frac{\pi\mu}{2}\right)+\frac{9}{8} \pi ^4 \tan ^2\left(\frac{\pi\mu}{2}\right) \sec^3\left(\frac{\pi\mu}{2}\right) +\frac{1}{16}\pi^{4}\tan^{4}\left(\frac{\pi\mu}{2}\right)\sec\left(\frac{\pi\mu}{2}\right)} \\[3mm]&={\pi \over 2}\,{5\pi^{4} \over 16} =\color{#00f}{\large{5\pi^{5} \over 32}} \end{align}

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  • $\begingroup$ Very helpful, thanks. This seems like a brute force method to solving these integrals for any power of the log $\int_0^{\pi/2} \log^n \tan \frac{x}{2}dx $, Thanks +1, EDit: $\Re(n)>-1/2$ for this to be convergent. $\endgroup$ – Jeff Faraci Apr 27 '14 at 4:58
  • $\begingroup$ @Integrals For $n$ odd it must be zero. $\endgroup$ – Felix Marin Apr 27 '14 at 4:59
  • $\begingroup$ Yes I have realized this. Thanks again. $\endgroup$ – Jeff Faraci Apr 27 '14 at 5:00
  • $\begingroup$ @FelixMarin. Why should it be zero if $n$ is odd ? I thought that, for any $n$, the result would express as $$(-1)^n 2^{-2 n-1} \left(\zeta \left(n+1,\frac{1}{4}\right)-\zeta \left(n+1,\frac{3}{4}\right)\right) \Gamma (n+1)$$ Where am I wrong ? Thanks. $\endgroup$ – Claude Leibovici Apr 27 '14 at 6:11
  • $\begingroup$ @ClaudeLeibovici Yes. You are right. It's the integral from $0$ to $\infty$ ( original from $0$ to $\pi$ ) which has that property. I was confused. Thanks for your remark. $\endgroup$ – Felix Marin Apr 27 '14 at 6:25
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This post and related comments and answers are very interesting and allow the generalization of the problem to $$I(n)=\int_0^{\pi/2}\log ^n\left[\tan \left(\frac{x}{2}\right)\right]dx$$ for which the result is $$I(n)=(-1)^n 2^{-2 n-1} \left[\zeta \left(n+1,\frac{1}{4}\right)-\zeta \left(n+1,\frac{3}{4}\right)\right] \Gamma (n+1)$$ When $n$ is even, one can find the "simple" formulas $$I(0)=\frac{\pi}{2}$$ $$I(2)=\frac{\pi ^3}{8}$$ $$I(4)=\frac{5 \pi ^5}{32}$$ $$I(6)=\frac{61 \pi ^7}{128}$$ $$I(8)=\frac{1385 \pi ^9}{512}$$ $$I(10)=\frac{50521 \pi ^{11}}{2048}$$ and so on.

When $n$ is odd, all values are negative and cannot be expressed in any form simpler than the $\zeta$ function.

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  • $\begingroup$ Thanks! Yes, The generalization of this problem is very nice. +1 $\endgroup$ – Jeff Faraci Apr 27 '14 at 16:02
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    $\begingroup$ Thank you for a good problem and you are very welcome. Cheers. $\endgroup$ – Claude Leibovici Apr 27 '14 at 16:04

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