Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need some help analytically proving the following with elementary tools:

$$\int_1^{+\infty} \frac{z^i + z^{-i}}{z^2 + 1} ~ \mathrm{d} z = \frac{\pi}{2} \mathrm{sech} \left ( \frac{\pi}{2} \right )$$

I tried substitution using the hyperbolic trigonometric functions, integrating by parts, but nothing seems to help make this integral simpler. I think the power of $i$ is what's giving me trouble, because if it was $z^3$ at the top for instance, it would be relatively easy. $z$ is a real, and the integral converges to a real as well.

I thought about multiplying the numerator and denominator by $z^i$ (and $z^{-i}$ for the other term) and using partial fractions to simplify the denominator into smaller terms I could integrate directly, but I am unsure how to proceed with exponents of $i$, since polynomials can only have real exponents. Wolfram Alpha doesn't simplify it either, which leads me to believe I can't do it that way.

share|cite|improve this question
Have you tried using the substitution z=e^x? See where that gets you! – user50165 Jan 1 '13 at 21:00
up vote 2 down vote accepted

Hint: expand in series $\dfrac{1}{z^2+1}$

share|cite|improve this answer
I don't see how exactly this could help, but anyway: wouldn't there be some problem as the series for the derivative of $\,\arctan z\,$ exists only for $\,|z|<1\,$ , whereas the integral is from one to $\,\infty\,$ ? – DonAntonio Jan 1 '13 at 11:42
Laurent series (i.e. series in powers of $1/z$), not Maclaurin series. – Robert Israel Jan 1 '13 at 21:13

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.