Solve the following definite integral: $\int_{0}^{\infty}\frac{x^2dx}{({1-x^2})^2}$

Question

Solve the following integral:

$$\int_{0}^{∞}\frac{x^2dx}{({1-x^2})^2}$$

I know that substituting some trigonometric functions may help. But I was not able to solve. Can you give me some clues?

Answer 1

Hint:

$\int_0^\infty\dfrac{x^2}{(1-x^2)^2}dx$

$=\int_0^1\dfrac{x^2}{(1-x^2)^2}dx+\int_1^\infty\dfrac{x^2}{(1-x^2)^2}dx$

$=\int_0^\infty\dfrac{\tanh^2x}{(1-\tanh^2x)^2}d(\tanh x)+\int_\infty^0\dfrac{\coth^2x}{(1-\coth^2x)^2}d(\coth x)$

$=\int_0^\infty\dfrac{\text{sech}^2x\tanh^2x}{\text{sech}^4x}dx+\int_0^\infty\dfrac{\text{csch}^2x\coth^2x}{\text{csch}^4x}dx$

$=\int_0^\infty(\sinh^2x+\cosh^2x)~dx$

$=\int_0^\infty\cosh2x~dx$

$=\left[\dfrac{\sinh2x}{2}\right]_0^\infty$

$=+\infty$

$\therefore\int_0^\infty\dfrac{x^2}{(1-x^2)^2}dx$ is divergent.

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\large\mbox{In case the}\ -\ \mbox{switch to}\ +$:

$$ \mu > 0\,, \quad \int_{0}^{\infty}{1 \over 1 + \mu x^{2}}\,{\rm d}x = {1 \over \sqrt{\mu\,}}\int_{0}^{\infty}{1 \over 1 + x^{2}}\,{\rm d}x = {1 \over \sqrt{\mu\,}}\,{\pi \over 2} $$

Derivate respect of $\ds{\mu}$ in both members: $$ -\int_{0}^{\infty}{x^{2} \over \left(1 + \mu x^{2}\right)^{2}}\,{\rm d}x = -\,{1 \over 2\mu^{3/2}}\,{\pi \over 2} $$

Set $\mu = 1$: $$ \int_{0}^{\infty}{x^{2} \over \left(1 + x^{2}\right)^{2}}\,{\rm d}x = {\pi \over 4} $$

$\large\mbox{With the}\ -\ \mbox{sign, we'll assume it's a 'principal value'}\ \pp $: \begin{align} \pp\int_{0}^{\infty}{x^{2} \over \left(1 - x^{2}\right)^{2}}\,{\rm d}x& =\lim_{\epsilon\ \to\ 0^{+}}\bracks{% \int_{0}^{1 - \epsilon}{x^{2} \over \left(1 - x^{2}\right)^{2}}\,{\rm d}x +\int_{1 + \epsilon}^{\infty}{x^{2} \over \left(1 - x^{2}\right)^{2}}\,{\rm d}x} \\[3mm]&=\lim_{\epsilon\ \to\ 0^{+}}\bracks{% \int_{0}^{1 - \epsilon}{x^{2} \over \left(1 - x^{2}\right)^{2}}\,{\rm d}x +\int_{1/\pars{1 + \epsilon}}^{0}{1/x^{2} \over \left(1 - 1/x^{2}\right)^{2}} \,\pars{-\,{\dd x \over x^{2}}}} \\[3mm]&=\lim_{\epsilon\ \to\ 0^{+}}\bracks{% \int_{0}^{1 - \epsilon}{x^{2} \over \left(1 - x^{2}\right)^{2}}\,{\rm d}x +\int_{0}^{1/\pars{1 + \epsilon}}{x^{2} \over \left(1 - x^{2}\right)^{2}} \,\dd x} \\[3mm]&=\lim_{\epsilon\ \to\ 0^{+}}\bracks{% 2\int_{0}^{1 - \epsilon}{x^{2} \over \left(1 - x^{2}\right)^{2}}\,{\rm d}x +\int_{1 - \epsilon}^{1/\pars{1 + \epsilon}} {x^{2} \over \left(1 - x^{2}\right)^{2}}\,\dd x} \end{align} The second integral $\ds{\stackrel{\epsilon\ \to\ 0^{+}}{\to}{1 \over 4}}$ while the first term behaves as $\ds{{1 \over 2\epsilon}}$ when $\ds{\epsilon \gtrsim 0}$ such that 'even' the "principal value" diverges.