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Compute the definite integral

$$ \int_0^\infty\frac{x^a-x^b}{(1+x^a)(1+x^b)}\,dx $$

where $a,b\in\mathbb{R}$.

My Attempt:

Let $x=\frac{1}{t}$ so that $dx=-\frac{1}{t^2}\,dt$. Substituting into the integral and changing the limits of integration gives

$$ \begin{align} \int_0^\infty\frac{x^a-x^b}{(1+x^a)(1+x^b)}\,dx&=\int_\infty^0\frac{t^b-t^a}{(t^a+1)(t^b+1)}\cdot\frac{-1}{t^2}\,dt\\ &=-\int_0^\infty\frac{t^a-t^b}{(1+t^a)(1+t^b)}\cdot\frac{1}{t^2}\,dt\\ &=-\int_0^\infty\frac{x^a-x^b}{(1+x^a)(1+x^b)}\cdot\frac{1}{x^2}\,dx \end{align} $$

I'm not sure how to compute the integral from here.

Fabrosi
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juantheron
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3 Answers3

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

Let's consider $\ds{\int_{0}^{\infty}{\dd x \over 1 + x^{\mu}}}$ with $\Re\pars{\mu} > 1$. With the change of variables $\ds{t \equiv {1 \over 1 + x^{\mu}}}$ $\iff$ $\ds{x = \pars{1 - t \over t}^{1/\mu}}$ \begin{align} \color{#00f}{\large\int_{0}^{\infty}{\dd x \over 1 + x^{\mu}}}&=\int_{1}^{0} t\,{1 \over \mu}\,\pars{1 - t \over t}^{1/\mu - 1}\,\pars{-\,{\dd t \over t^{2}}} ={1 \over \mu}\int_{0}^{1}t^{-1/\mu}\pars{1 - t}^{1/\mu - 1}\,\dd t \\[3mm]&={1 \over \mu}\,{\rm B}\pars{-\,{1 \over \mu} + 1,{1 \over \mu}} ={1 \over \mu}\, {\Gamma\pars{-1/\mu + 1}\Gamma\pars{1/\mu} \over \Gamma\pars{\bracks{-1/\mu + 1} + 1/\mu}} ={1 \over \mu}\,{\pi \over \sin\pars{\pi\,\bracks{1/\mu}}} \\[3mm]&=\color{#00f}{\large{\pi \over \mu}\,\csc\pars{\pi \over \mu}} \end{align}

Then, $$\!\!\!\color{#00f}{\large% \int_{0}^{\infty}\!\!\!{x^{a} - x^{b} \over \pars{1 + x^{a}}\pars{1 + x^{b}}}\,\dd x = {\pi \over b}\,\csc\pars{\pi \over b} - {\pi \over a}\,\csc\pars{\pi \over a}} \,,\qquad\Re\pars{a} > 1\,,\ \Re\pars{b} > 1 $$

Felix Marin
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I must say that I am embarassed to give an answer ignoring what is your knowledge in the area of special functions. So, please, forgive me is this is out of your scope.

The antiderivative $$I=\int \frac{dx}{1+x^a}=x \, _2F_1\left(1,\frac{1}{a};1+\frac{1}{a};-x^a\right)$$where appears the hypergeometric function. Concerning the integral $$I=\int_0^\infty \frac{dx}{1+x^a}= \frac{\pi }{a}\, \csc \left(\frac{\pi }{a}\right)$$ provided $\Re(a)>1$.

Claude Leibovici
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If you use partial fractions, you will see that the integrand is $$\frac{1}{1+x^b} - \frac{1}{1+x^a}$$ Neither summand can be indefinitely integrated in elementary terms, but the residue theorem is your friend.

Igor Rivin
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  • Thanks Igor Rivin, would you like to explain me the residue theorem, bcz i did not have a knowledge of Residue Theorem, Thanks – juantheron Dec 12 '13 at 03:45
  • @Igor Rivin. If the OP does not know abour the residue theorem, how do you suggest to compute the integral ? I know the answer to his problem but I wonde how to explain the way to proceed. Thanks and cheers. – Claude Leibovici Dec 12 '13 at 05:03
  • @ClaudeLeibovici Well, I am a little stumped, though I do have some ideas. How would you do it? – Igor Rivin Dec 12 '13 at 05:12
  • @IgorRivin. That is the good question ! The antiderivative involves an hypergeometric 2F1 function. The result of the integral itself is quite simple. But I really do not know what to suggest for the integration. – Claude Leibovici Dec 12 '13 at 05:28
  • @ClaudeLeibovici there are often tricks, involving differentiating/integrating with respect to parameter, or taking Laplace transform and limit as $s\rightarrow 0,$ but for some reason I am not seeing anything here. – Igor Rivin Dec 12 '13 at 05:35
  • After you have written it using partial fractions, see here to show that the final result is $$ \frac{\pi}{b}\csc\left(\frac{\pi}{b}\right) - \frac{\pi}{a}\csc\left(\frac{\pi}{a}\right). $$ – Bennett Gardiner Dec 12 '13 at 23:52