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In my homework there is a integral looks so strange, this is the first time that I see this kind of thing. So I have no idea what to do.

$\int_{0}^{-1+i\infty}(z+1)e^{iz}dz$

What does it mean the upper part of integral $-1+i \infty$? And how can i solve it?

I tried to take integral respect to $z$. Then i wrote $z=x+iy$ and put from $0$ to $-1$ for all x and put from $0$ to $\infty$ for all y. Am I right?

Aegean
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  • Observe that the function $(z+1)e^{iz}$ is analytic, so that the integral does not depend on a path. The simplest way is to use the antiderivative. – user Mar 26 '21 at 10:37

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It is better to consider first $$I_p=\int_{0}^{-1+ip}(z+1)\,e^{iz}\,dz$$ Expandin the integrand, one integration by parts gives the antiderivative and using the bounds $$I_p=e^{-(p+i)} (p+1)-(1-i)$$ Now, take the limit when $p\to \infty$.

Claude Leibovici
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  • when p goes to $\infty$, $e^{-\infty}=0$ so left side is $0$ and $i-1$ is the answer, right? I found exactly the same with the method I said – Aegean Mar 26 '21 at 12:04