Take the 2-minute tour ×
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.

I am looking to evaluate the following improper-integral

$$ \int_0^\infty \left( \frac{1}{ 1 + x^\alpha \vert v-1 \vert^\alpha} - \frac{1}{ 1 + x^\alpha \vert v+1 \vert^\alpha} \right) \frac{ d v}{v} $$

for some positive value of $x$ and $\alpha = 3$ in view of being able to generalize this to odd integers $\alpha = 2 n-1$.

The answer for $\alpha=3$ is known to be

$$ \frac{2 x^3 \log(1/x)}{1-x^6} + \frac{\pi x s_3}{ (x-c_3)^2 + s_3^2} + \frac{\pi x s_3}{ (x+c_3)^2 + s_3^2} $$

where $s_3 = \sin\left(\pi/3\right)$ and $c_3 = \cos\left(\pi/3\right)$.

I am not able to reproduce this, and would appreciate help. Thank you

share|improve this question

1 Answer 1

up vote 7 down vote accepted

An integral from $0$ to $\infty$ with $\mathrm dv/v$ in it tends to be susceptible to a substitution of the form $u=\lambda v$, which leaves all of that invariant. In the present case, you can write $u=xv$ to get

$$ \int_0^\infty \left( \frac{1}{ 1 + x^\alpha \vert v-1 \vert^\alpha} - \frac{1}{ 1 + x^\alpha \vert v+1 \vert^\alpha} \right) \frac{\mathrm d v}{v} = \int_0^\infty \left( \frac{1}{ 1 + \vert u-x \vert^\alpha} - \frac{1}{ 1 + \vert u+x \vert^\alpha} \right) \frac{\mathrm d u}{u}\;. $$

The integrand is even, so you can simplify things by calculating the integral from $-\infty$ to $\infty$ instead.

The denominators are of the form $z^\alpha+1$, which for odd $\alpha$ factorizes into $\prod_i(z+z_i)$, where $z_i$ are the $\alpha$-th roots of unity. So we have

$$\frac12\int_{-\infty}^\infty \left( \frac{1}{\prod_i(\vert u-x \vert+z_i)} - \frac{1}{\prod_i(\vert u+x \vert+z_i)} \right) \frac{\mathrm d u}{u}\;.$$

The rest is an exercise in partial fractions, with the two complex conjugate solutions combining into a quadratic denominator; I think you'll need to resolve the absolute values first; let me know if you want me to write it out further.

P.S.: You can further simplify things by substituting $t=u-x$ in the first term and $t=u+x$ in the second; that yields

$$\frac12\int_{-\infty}^\infty \frac{1}{\prod_i(\vert t \vert+z_i)}\left(\frac1{t+x}-\frac1{t-x} \right) \mathrm d t\;. $$

share|improve this answer
Very nice solution! –  JavaMan Aug 2 '11 at 21:23
@joriki Thank you for a clear explanation. The last integral should be treated as principal value integral. –  Sasha Aug 2 '11 at 23:23
@Sasha: Yes, I missed that. And you're welcome. –  joriki Aug 3 '11 at 3:53

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.