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I am trying to prove this interesting integral$$ \mathcal{I}:=\int_0^{\pi/4}\frac{dx}{{\sin 2x}\,(\tan^ax+\cot^ax)}=\frac{\pi}{8a},\qquad \mathcal{Re}(a)\neq 0. $$ This result is breath taking but I am more stumped than usual. It truly is magnificent. I am not sure how to approach this, note $\sin 2x=2\sin x \cos x$. I am not sure how to approach this because of the term $$ (\tan^ax+\cot^ax) $$ in the denominator. I was trying to use the identity $$ \tan \left(\frac{\pi}{2}-x\right)=\cot x $$ since this method solves a similar kind of integral but didn't get anywhere. A bad idea I tried was to try and differentiate with respect to a $$ \frac{dI(a)}{da}=\int_0^{\pi/4}\partial_a \left(\frac{dx}{{\sin 2x}\,(\tan^ax+\cot^ax)}\right)=\int_0^{\pi/4} \frac{(\cot^a x \log(\cot x )+\log(\tan x ) \tan^a x)}{\sin 2x \, (\cot^a x+\tan^a x)^2}dx $$ which seems more complicated when I break it up into two integrals. How can we solve the integral? Thanks.

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  • $\begingroup$ This one just begs for a Weierstrass substitution. Also, $\cot=\dfrac1\tan$ $\endgroup$ – Lucian May 23 '14 at 4:23
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Rewrite: \begin{align} \int_0^{\Large\frac\pi4}\frac{dx}{{\sin 2x}\,(\tan^ax+\cot^ax)}&=\int_0^{\Large\frac\pi4}\frac{dx}{{2\sin x\cos x}\,\left(\tan^ax+\dfrac1{\tan^ax}\right)}\\ &=\frac12\int_0^{\Large\frac\pi4}\frac{\tan^ax\ dx}{{\tan x\cos^2x}\,\left(1+\tan^{2a}x\right)}\\ &=\frac12\int_0^{\Large\frac\pi4}\frac{\tan^{a-1}x\ d(\tan x)}{1+\tan^{2a}x}. \end{align} Now, let $\tan^a x=\tan\theta\;\Rightarrow\;a\tan^{a-1}x\ d(\tan x)=\sec^2\theta\ d\theta$. Then \begin{align} \frac12\int_{x=0}^{\Large\frac\pi4}\frac{\tan^{a-1}x\ d(\tan x)}{1+\tan^{2a}x}&=\frac1{2|a|}\int_{\theta=0}^{\Large\frac\pi4}\frac{\sec^2\theta\ d\theta}{1+\tan^{2}\theta}\\ &=\frac1{2|a|}\int_{\theta=0}^{\Large\frac\pi4}\ d\theta\\ &=\large\color{blue}{\frac{\pi}{8|a|}}. \end{align}

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  • 1
    $\begingroup$ @Integrals Jeff, why did you suspect your account would be suspended? Is it wrong if you ask lots of integral questions? I don't think so considering you're not a student which is fine to me asking too much as a curiousity. I have to admit, I learn a lot from your integral questions. So, I really hope the moderator will not suspend your account and keep posting integral questions! :) $\endgroup$ – Tunk-Fey May 23 '14 at 9:07
  • $\begingroup$ I made too many edits and posted too many problems, therefore flooded the main page. I didn't know this, I just post integrals because this is a math site. However the message was a "warning". People seem to be bothered by a lot of integrals, I have a lot of detail in every post so I am not sure what the issue is. $\endgroup$ – Jeff Faraci May 24 '14 at 19:36
  • $\begingroup$ by the way +1 This solution is excellent and complete! $\endgroup$ – Jeff Faraci May 24 '14 at 19:38
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    $\begingroup$ @Integrals There are some users that believe they are a sort of M.SE police officer. One of them make campaign to delete one of my answers. That answer took me several hours. It was a detailed solution of the $2D$-Laplace equation where I found a closed solutions in terms of $\left(r,\theta\right)\ \mbox{and}\ \left(x,y\right)$. It was something like a $2D$-Electrostatic problem. Other solutions were hints and answers with series. This user even criticizes how I write LaTeX. It's always around acting as a soldier. Unfortunately, I didn't keep a copy of it and it was lost forever. $\endgroup$ – Felix Marin Jul 22 '14 at 4:36
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    $\begingroup$ @Integrals I just see your new post. It's true: It looks like a 2-$D$ electrostatic problem. I'll think about it. Thanks. $\endgroup$ – Felix Marin Aug 31 '14 at 20:21
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{{\cal I}\equiv\int_{0}^{\pi/4}{\dd x \over \sin\pars{2x}\bracks{\tan^{a}\pars{x} + \cot^{a}\pars{x}}}={\pi \over 8\verts{a}},\qquad \Re\pars{a} \neq 0}$.

Note that the integral depends obviously on $\ds{\large\verts{a}}$.

\begin{align} {\cal I}&=\half\ \overbrace{\int_{0}^{\pi/2}{\dd x \over \sin\pars{x} \bracks{\tan^{\verts{a}}\pars{x/2} + \cot^{\verts{a}}\pars{x/2}}}} ^{\ds{t \equiv \tan\pars{x \over 2}}} \\[3mm]&=\half\int_{0}^{1}{2\,\dd t/\pars{1 + t^{2}} \over \bracks{2t/\pars{1 + t^{2}}}\pars{t^{\verts{a}} + t^{-\verts{a}}}} =\half\ \overbrace{\int_{0}^{1} {t^{\verts{a} - 1} \over t^{2\verts{a}} + 1}\,\dd t} ^{\ds{t^{\verts{a}} \equiv x}\ \imp\ t = x^{1/\verts{a}}} \\[3mm]&=\half\int_{0}^{1}{ \pars{x^{1/\verts{a}}}^{\verts{a} - 1} \over x^{2} + 1}\,{1 \over \verts{a}}\,x^{1/\verts{a} - 1}\dd x ={1 \over 2\verts{a}}\ \overbrace{\int_{0}^{1}{\dd x \over x^{2} + 1}} ^{\ds{=\ {\pi \over 4}}} \end{align}

$$\color{#00f}{\large% {\cal I}\equiv\int_{0}^{\pi/4}{\dd x \over \sin\pars{2x}\bracks{\tan^{a}\pars{x} + \cot^{a}\pars{x}}}={\pi \over 8\verts{a}}}\,,\qquad\Re\pars{a} \not= 0 $$

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  • $\begingroup$ I fix my answer, thanks for your concern Sir. Anyway +1. :) $\endgroup$ – Tunk-Fey May 25 '14 at 4:03
  • $\begingroup$ @Tunk-Fey Fine. Thanks a lot. $\endgroup$ – Felix Marin May 25 '14 at 5:35
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Start with $\tan x=t$ and then $t^a=\tan u$. You could have used them together but this looks better. All terms beautifully(as planned), cancel, giving : $$\int_0^{\dfrac {\pi}{4}}\dfrac{1}{2a}du$$

I have no idea how to do this one though.

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