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

How to determine where $$f(z)=\int_0^\infty \frac{e^{tz}}{1+t^2} \, dt$$ is defined and holomorphic using Morera's and Fubini's theorem?

share|cite|improve this question
So, what would you need to do to apply Morera's theorem to show that $f(z)$ is analytic in a region $G$? – GEdgar Sep 24 '12 at 1:16
@MhenniBenghorbal: what happens if limits of integration is from -1 to 1? – abby Sep 24 '12 at 1:54
@abby:The integral has no problem at these points. See the answer. – Mhenni Benghorbal Sep 24 '12 at 11:35

Related problems: (I), (II). The region of convergence of the integral is $ Re(z) \leq 0 \,.$ To prove $f(z)$ is analytic, we appeal to Morera's theorem which states that a continuous, complex-valued function ƒ defined on a connected open set $D$ (in your case $D=\{z: Re(z)< 0 \})$ in the complex plane that satisfies $$ \oint_{\gamma} f(z)\, dz = 0 $$ for every closed piecewise $C_1$ curve $γ$ in $D$ must be holomorphic on $D$.

Applying the theorem to your case, we have

$$ \oint_{\gamma} f(z) dz = \oint \int_{0}^{\infty} \frac{{\rm e}^{zt}}{t^2+1}dt dz = \int_{0}^{\infty}\frac{1}{t^2+1}\oint_{\gamma} {\rm e}^{zt} dz \,dt = \int_{0}^{1} \frac{1}{t^2+1} (0) dt = 0 \,.$$

The inner integral equals 0, since ${\rm e}^{zt}$ is analytic and hence by Cauchy theorem the integral is zero. The interchanging of the integrals is justified by the uniform convergence of the $\int_0^\infty \frac{e^{tz}}{1+t^2} \, dt $ or you can just apply Fubini's theorem.

share|cite|improve this answer

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.