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What are the steps to evaluate the following definite integral? (Answer provided)

$$\int_0^\infty {{\pi^2\over 4}\cos^2x\over\left({\pi^2\over4}-x^2 \right)^2} dx={\pi\over 4}$$?

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2 Answers 2

Contour integration is how I would approach this integral. $$ \begin{align} \int_0^\infty {{\pi^2\over 4}\cos^2x\over\left({\pi^2\over4}-x^2 \right)^2}\;\mathrm{d}x &=\frac{1}{2}\int_{-\infty}^\infty {{\pi^2\over 4}\cos^2x\over\left({\pi^2\over4}-x^2 \right)^2} \;\mathrm{d}x\tag{1}\\ &=\frac{\pi^2}{2}\int_{-\infty}^\infty\frac{e^{2ix}+2+e^{-2ix}}{(\pi^2-4x^2)^2}\mathrm{d}x\tag{2}\\ &=\frac{\pi^2}{2}\int_{-\infty-i\epsilon}^{\infty-i\epsilon}\frac{e^{2ix}+2+e^{-2ix}}{(\pi^2-4x^2)^2}\mathrm{d}x\tag{3}\\ &=\frac{\pi^2}{2}\oint_{\gamma^+}\frac{e^{2ix}+1}{(\pi^2-4x^2)^2}\mathrm{d}x\\ &+\frac{\pi^2}{2}\oint_{\gamma^-}\frac{e^{-2ix}+1}{(\pi^2-4x^2)^2}\mathrm{d}x\tag{4}\\ &=\frac{\pi^2}{2}\oint_{\gamma^+}\frac{e^{2ix}+1}{(\pi^2-4x^2)^2}\mathrm{d}x\tag{5}\\ \end{align} $$

  1. make the path a bit nicer. Since the integrand is even, let's duplicate the domain of integration and divide by two

  2. expand $\cos^2(x)=\dfrac{e^{2ix}+2+e^{-2ix}}{4}$

  3. move the path of integration. We cross no non-removable singularities and the connections near $+\infty$ and $-\infty$ vanish.

  4. break up the integral into two contours: $\gamma^+$ which passes from $-\infty-i\epsilon$ to $+\infty-i\epsilon$ and then circles back counter-clockwise around the upper half-plane, and $\gamma^-$ which passes from $-\infty-i\epsilon$ to $+\infty-i\epsilon$ and then circles back clockwise around the lower half-plane.

  5. $\gamma^-$ circles no singularities, so its integral is $0$.

Account for the residues of the singularities in $(5)$. $$ \small\begin{array}{c} \text{singularity}&&\text{integrand}&&\text{first order}&&\text{residue}\\ x=\frac{\pi}{2}&:&\frac{1-e^{2i(x-\pi/2)}}{(-2(x-\pi/2)(2\pi+2(x-\pi/2)))^2}&\to&\frac{-2i(x-\pi/2)}{16\pi^2(x-\pi/2)^2}&\to&\frac{1}{4\pi}\\ x=-\frac{\pi}{2}&:&\frac{1-e^{2i(x+\pi/2)}}{(2(x+\pi/2)(2\pi-2(x+\pi/2)))^2}&\to&\frac{-2i(x+\pi/2)}{16\pi^2(x+\pi/2)^2}&\to&\frac{1}{4\pi} \end{array} $$ Thus, the sum of the residues is $\frac{1}{2\pi}$. Including the $\frac{\pi^2}{2}$ yields an integral of $\frac{\pi}{4}$.

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{\infty}{{\pi^{2} \over 4}\cos^{2}\pars{x} \over \pars{{\pi^{2} \over 4} -x^{2}}^{2}}\,\dd x = {\pi \over 4}:\ {\large ?}}$

\begin{align} \pp\int_{-\infty}^{\infty}{{\pi^{2} \over 4}\cos^{2}\pars{x} \over x^{2} - \mu} \,\dd x&= {\pi^{2} \over 8\root{\mu}}\,\pp\int_{-\infty}^{\infty} \bracks{{\cos^{2}\pars{x} \over x - \root{\mu}} - {\cos^{2}\pars{x} \over x + \root{\mu}}}\,\dd x \\[3mm]&= {\pi^{2} \over 8\root{\mu}}\,\pp\int_{-\infty}^{\infty} {\cos^{2}\pars{x + \root{\mu}} - \cos^{2}\pars{x - \root{\mu}} \over x}\,\dd x \\[3mm]&= {\pi^{2} \over 8\root{\mu}}\,\pp\int_{-\infty}^{\infty} {-4\cos\pars{x}\cos\pars{\root{\mu}}\sin\pars{x}\sin\pars{\root{\mu}} \over x}\,\dd x \\[3mm]&= -\,{\pi^{2}\sin\pars{2\root{\mu}} \over 8\root{\mu}}\,\pp\int_{-\infty}^{\infty} {\sin\pars{2x} \over x}\,\dd x = -\,{\pi^{3} \over 8}\,{\sin\pars{2\root{\mu}} \over \root{\mu}} \end{align}

\begin{align} \pp\int_{-\infty}^{\infty} {{\pi^{2} \over 4}\cos^{2}\pars{x} \over \pars{x^{2} - \mu}^{2}} \,\dd x&= -\,{\pi^{3} \over 8}\, \totald{}{\mu}\bracks{{\sin\pars{2\root{\mu}} \over \root{\mu}}} = -\,{\pi^{3} \over 8}\, \bracks{{\cos\pars{2\root{\mu}} \over \mu} - {\sin\pars{2\root{\mu}} \over 2\mu^{3/2}}} \end{align}

Set $\mu = \pi^{2}/4$: \begin{align} \int_{-\infty}^{\infty} {{\pi^{2} \over 4}\cos^{2}\pars{x} \over \pars{x^{2} - \pi^{2}/4}^{2}} \,\dd x&= -\,{\pi^{3} \over 8}\,\pars{-1 \over \pi^{2}/4} = {\pi \over 2} \end{align}

$$ \color{#00f}{\large\int_{0}^{\infty}{{\pi^{2} \over 4}\cos^{2}\pars{x} \over \pars{{\pi^{2} \over 4} -x^{2}}^{2}}\,\dd x} = \half\int_{-\infty}^{\infty} {{\pi^{2} \over 4}\cos^{2}\pars{x} \over \pars{x^{2} - \pi^{2}/4}^{2}} \,\dd x = \color{#00f}{\large{\pi \over 4}} $$

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