Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question

3 Answers 3

up vote 10 down vote accepted

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

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

share|cite|improve this answer

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 $\Re(a)>1$.

share|cite|improve this answer
LaTeX, please... – vonbrand Aug 24 at 23:22
@vonbrand. It took me a very long time to use LaTex since I am almost blind. Now, there are things I can write almost correctly but my old answers still suffer that lack. Cheers. – Claude Leibovici Aug 25 at 5:57

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.

share|cite|improve this answer
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 3:45
@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 5:03
@ClaudeLeibovici Well, I am a little stumped, though I do have some ideas. How would you do it? – Igor Rivin Dec 12 '13 at 5:12
@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 5:28
@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 5:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.