What's a straightforward proof that $$\int_0^1 e^{zt}dt$$ ($z \in \mathbb{C}$) is holomorphic in $\mathbb{C}$? Also, how can we calculate the integral?
Holomorphicity and value of $\int_0^1 e^{zt}dt$
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integration
complex-analysis
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0Integrate with respect to $t$, holding $z$ 'constant' ending with $$\frac{e^{zt}}{z}\biggr\rvert_0^1$$. Then apply boon het's answer below. – 2017-01-06
1 Answers
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This function is nothing but $$\begin{cases}\frac{e^z-1}{z} & z\neq 0 \\ 1 & z=0\end{cases}$$ which is clearly holomorphic away from zero. Check its holomorphicity at zero and you will be done.