# Find the region where an integral is defined

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?

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