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We need to find out the domain for which the following function is defined and holomorphic:

$$f(z)=\int\limits_{-1}^{1}\frac{e^{tz}}{1+t^2}dt$$

How do we proceed to this kind of problem?

Any hints please?

gamma
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  • The first thing to do is to make some kind of guess at which values of $z$ will work. Which ones do you think the integral will converge for? – Antonio Vargas Nov 25 '15 at 11:13
  • There was a mistake in the question. Sorry. – gamma Nov 25 '15 at 11:26
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    Hint: $$f(z)=\sum_{n=0}^\infty a_n\frac{z^n}{n!}\qquad a_n=\int_{-1}^1\ldots dt,$$ hence, for every $n$, $$|a_n|\leqslant\ldots$$ which proves that the radius of convergence is $_____$. – Did Nov 25 '15 at 12:22
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    Related: http://math.stackexchange.com/questions/179088/showing-a-function-defined-from-an-integral-is-entire – ChaPi Nov 25 '15 at 12:24

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Morera's theorem is probably the simplest approach (for questions such as these -- Did's hint gives a quicker solution in this particular case). First show that $f$ is continuous. (I'll leave that to you.) Then, if $\gamma$ is any closed curve in $\mathbb{C}$, \begin{align} \int_{\gamma} f(z)\,dz &= \int_{\gamma} \left( \int_0^1 \frac{e^{tz}}{1+t^2}\,dt \right) \,dz \\ &= \int_{-1}^1 \left( \int_{\gamma} \frac{e^{tz}}{1+t^2}\,dz \right) \,dt \\ &= \int_{-1}^1 \left( \int_{\gamma} e^{tz} \,dz \right) \frac{1}{1+t^2} \,dt = \int_{-1}^1 0\,dt = 0\\ \end{align} where the inner integral (after changing the order of integration) vanishes due to Cauchy's integral theorem. Morera's theorem then shows that $f$ is holomorphic everywhere.

mrf
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