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I have tried solving this for about an hour and will probably resort to head banging in some time:

$$\int ^{\frac{\pi}{2}}_{0} \dfrac{dx}{(a^2\cos^2 x + b^2 \sin ^2 x)^2}$$

I first divided by $\cos^4 x$ and then subsequently put $\tan x = t$, to get:

$$\int ^{\infty}_{0} \dfrac{1+t^2}{(a^2 + b^2 t^2)^2}dt$$

This has become unmanageable. Neither splitting the numerator, nor Partial fraction (taking t^2 = z and applying partial fraction) seems to work.

Did
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jonsno
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5 Answers5

14

For a solution that doesn't involve complex analysis: Try the substitution $t = \frac{a}{b} \tan u$. Then $du = \frac{ab}{a^2 + b^2 t^2} \,dt$ and your integral becomes $$\frac{1}{a^3 b^3} \int_0^{\pi/2} (b^2 \cos^2 u + a^2 \sin^2 u) \,du$$ which is more doable.

arkeet
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7

Say that $ a, b >0 .$ From the substitution $t \mapsto \dfrac{a}{b}t$ followed by $ t \mapsto \dfrac{1}{t}$ we get $$ I = \int_{0}^{\infty}\dfrac{1+t^{2}}{(a^{2}+b^{2}t^{2})^{2}}\, dt = \dfrac{1}{a^{3}b^{3}} \int_{0}^{\infty}\dfrac{b^2+a^2t^{2}}{(1+ t^{2})^{2}}\, dt $$ and $$ I = \dfrac{1}{a^{3}b^{3}} \int_{0}^{\infty}\dfrac{b^2t^2+a^2}{(1+ t^{2})^{2}}\, dt . $$ Consequently $$ 2I = \dfrac{a^2+b^2}{a^3b^3}\int_{0}^{\infty}\dfrac{1}{1+t^2}\, dt = \dfrac{\pi}{2}\dfrac{a^2+b^2}{a^3b^3} $$ and $$ I = \dfrac{\pi}{4}\dfrac{a^2+b^2}{a^3b^3}. $$

JanG
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5

$f(z)=\frac{1+z^2}{(a^2+b^2 z^2)^2}$ is an even meromorphic function with double poles at $z=\pm\frac{a}{b}i$, that decays like $\frac{1}{|z|^2}$ as $|z|\to +\infty$. It follows that your integral equals

$$\pi i\cdot\text{Res}\left(f(z),z=\frac{a}{b}i\right)=\color{red}{\frac{\pi(a^2+b^2)}{4a^3b^3}} $$ due to the residue theorem and the ML lemma. For practicing headbanging, this is better.

Jack D'Aurizio
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  • Thank you a lot but I do not have even the slightest idea of Res and meromorphic function and its decay! – jonsno Sep 24 '16 at 01:58
  • @samjoe: it may be a good moment for reading the linked material, since it is pretty useful for dealing with such kind of integrals. – Jack D'Aurizio Sep 24 '16 at 02:04
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$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\color{#f00}{\int_{0}^{\infty} {1 + t^{2} \over \pars{a^{2} + b^{2}t^{2}}^{2}}\,\dd t} = {1 \over \verts{b}^{3}}\int_{0}^{\infty} {b^{2} + t^{2} \over \pars{a^{2} + t^{2}}^{2}}\,\dd t \\[5mm] = &\ {1 \over \verts{b}^{3}}\bracks{\pars{b^{2} - a^{2}}\int_{0}^{\infty} {\dd t \over \pars{t^{2} + a^{2}}^{2}} + \int_{0}^{\infty}{\dd t \over t^{2} + a^{2}}} \\[5mm] = &\ {1 \over \verts{b}^{3}}\bracks{\pars{a^{2} - b^{2}}\partiald{}{{a^{2}}} + 1} \int_{0}^{\infty}{\dd t \over t^{2} + a^{2}} = {1 \over \verts{b}^{3}}\bracks{\pars{a^{2} - b^{2}}\partiald{}{{a^{2}}} + 1} {\pi \over 2}\pars{a^{2}}^{-1/2} \\[5mm] = &\ {\pi \over 2\verts{b}^{3}}\braces{\pars{a^{2} - b^{2}} \bracks{-\,{1 \over 2}\pars{a^{2}}^{-3/2}} + \pars{a^{2}}^{-1/2}} = {\pi \over 4\verts{a}^{3}\verts{b}^{3}}\pars{-a^{2} + b^{2} + 2a^{2}} \\[5mm] = &\ \color{#f00}{{\pi \over 4}\,{a^{2} + b^{2} \over \verts{a}^{3}\verts{b}^{3}}} \end{align}

Felix Marin
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  • can we apply leibniz rule to constant? – jonsno Oct 01 '16 at 15:12
  • @samjoe Thanks for your remark. ${\partial,\mathrm{f}\left(a^{2}\right) \over \partial a^{2}}$ is a short-cut for $\left. {\partial,\mathrm{f}\left(p\right) \over \partial p}\right\vert_{\ p\ =\ a^{2}}$. It just avoids a 'visually cumbersome' longer line. That's all. You can rewrite everything without the short-cut. No problem. But the short-cut improves readability. – Felix Marin Oct 01 '16 at 22:11
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$$ \begin{align} \int_0^\infty\frac{1+t^2}{(a^2+b^2t^2)^2}\,\mathrm{d}t &=\frac1{a^3b}\int_0^\infty\frac{1+\frac{a^2}{b^2}t^2}{(1+t^2)^2}\,\mathrm{d}t\tag{1}\\ &=\frac1{2a^3b}\int_0^\infty\frac{t^{-1/2}+\frac{a^2}{b^2}t^{1/2}}{(1+t)^2}\,\mathrm{d}t\tag{2}\\ &=\frac1{2a^3b}\frac{\Gamma\left(\frac12\right)\Gamma\left(\frac32\right)}{\Gamma(2)}+\frac1{2ab^3}\frac{\Gamma\left(\frac32\right)\Gamma\left(\frac12\right)}{\Gamma(2)}\tag{3}\\[3pt] &=\frac\pi4\frac{a^2+b^2}{a^3b^3}\tag{4} \end{align} $$ Explanation:
$(1)$: substitute $t\mapsto\frac abt$
$(2)$: substitute $t\mapsto t^{1/2}$
$(3)$: use the Beta Function integral
$(4)$: simplify

robjohn
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