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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?

  • 0
    So, what would you need to do to apply Morera's theorem to show that $f(z)$ is analytic in a region $G$?2012-09-24
  • 0
    @MhenniBenghorbal: what happens if limits of integration is from -1 to 1?2012-09-24
  • 0
    @abby:The integral has no problem at these points. See the answer.2012-09-24

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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.