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$$\frac{1}{2\pi i}\int_{I(\lambda,\infty)}^{} \left(\frac{1}{t}+\frac{1}{2}+\frac{1}{12}t\right)\frac{e^{-zt}\log t}{t^2}\,dt$$ where $I(\lambda, \infty)$ is the integral path consisting of $(\infty, \lambda)$ , counterclockwise circle of radius $\lambda$ around the origin and $(\lambda, +\infty)$

Please teach me how to calculate this contour integrate.

O-V-E-R-H-E-A-T
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  • For $\Re(s) > 0$, $\int_{I(\lambda,\infty)} t^{s-1} e^{-t}dt =\lim_{\lambda \to 0} \int_{I(\lambda,\infty)} t^{s-1} e^{-t}dt= \int_\infty^0 + \int_0^{e^{2i \pi} \infty} t^{s-1} e^{-t}dt$ $ = (e^{2i \pi (s-1)}-1) \int_0^\infty t^{s-1} e^{-t}dt = (e^{2i \pi s}-1) \Gamma(s)$. Then the result stays true for every $s$ by analytic continuation. Do you see how it helps for your question when $z = 1$ and $z > 0$, and for $\Re(z) > 0$ by analytic continuation ? – reuns Dec 21 '17 at 10:26
  • Is it possible you meant "counterclockwise half circle of radius..." ? – DonAntonio Dec 21 '17 at 10:27
  • @DonAntonio No. The contour is thisone without the large circle. – reuns Dec 21 '17 at 10:34
  • @reuns I understood it. But I have no idea how to use it for my question...sorry – O-V-E-R-H-E-A-T Dec 21 '17 at 10:59
  • @reuns How can you tell? Anyway, it looks like that indeed is the contour, a branched one for the logarithm. – DonAntonio Dec 21 '17 at 11:01
  • Look at $\frac{d}{ds} \int_{I(\lambda,\infty)} t^{s-1} e^{-t}dt$ to make $\log t$ appear. @DonAntonio Because this contour is well-known for functions with a single branch point at $0$. – reuns Dec 21 '17 at 11:10
  • @reuns Yes, I know...yet the option to take away the non-negative real axis was what prompted me to ask for a clarification, as the usual branch cut for the logarithm is the non-positive axis...but it is logical. Thanks. – DonAntonio Dec 21 '17 at 11:12

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