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Here's my problem: I have this function $\psi(\lambda)=\frac{e}{\pi\lambda}Im(e^{-\omega(\lambda-1)^{\frac{1}{4}}})$ (where $\omega=e^{-i\frac{\pi}{4}}$) defined on the interval $[1,\infty]$ and want to compute its integral (it should be 1). I've been suggested to consider the complex function $e^{-\omega(z-1)^{\frac{1}{4}}}$ (cutting from the complex plane the interval $[1,\infty]$). Integrating on the curve $\gamma$ that goes from $\infty$ to 1, just above the slit, then doing a half-loop around 1 and going back to $\infty$ under the cut, I easily obtain from the residue formula that $-\frac{1}{2\pi i}\int_{\gamma}\frac{1}{z}e^{-\omega(z-1)^{\frac{1}{4}}}dz=e^{-1}$

Now, I guess, I have to take the limit (taking $\gamma$ closer and closer to $[1,\infty]$). However this part is a little bit tricky and I don't have figured out yet how to proceed.


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What is $\omega$? – badp Jan 16 '11 at 11:02
Sorry, $\omega=e^{-i\frac{\pi}{4}}$. – Gabrio Jan 16 '11 at 11:48

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