How to compute $\displaystyle\int_0^\infty \frac{\sin t}{t^{s+1}}\;\text dt$ ?
Here, the real part of the complex number $s$ is negative and greater than $-1$.
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What have you tried? Computing $\displaystyle\int_0^\infty \cfrac{e^{\varepsilon it}}{t^{s-1}}dt$ for $\varepsilon \in \left{-1,1\right}$? Series? – xavierm02 May 24 '13 at 16:36
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I have run Mathematica to obtain $\int_0^\infty \frac{\sin t}{t^{s+1}} dt=-\Gamma(-s) \sin \frac{s\pi}2$, and tried computing $\int_0^\infty \frac{e^{it}}{t^{s+1}}dt$ as we do when computing $\int_0^\infty \frac{\sin t}{t} dt$. But I lost the way. – David John May 24 '13 at 17:05
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Integrate $\frac{e^{iz}}{z^{s+1}}$ around a closed quarter-circle in the first quadrant of the complex plane that is indented at the origin. And $- 1 < \text{Re}(s) < 1$. – Random Variable May 24 '13 at 17:22
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Using the definition of $\Gamma$-function from here, prove: $$\int_0^\infty e^{-a t}\,t^{s-1}\,dt=\Gamma(s)\,a^{-s},$$ then represent $$\sin t=\frac{e^{i t}-e^{-i t}}{2\,i}.$$
OlegK
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Many thanks! Does the above equality hold even when $a$ is a complex number? – David John May 24 '13 at 17:30
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In general it holds for $\text{Re}(a) >0$. But integrating around the contour I suggested above will also show it holds for $a$ purely imaginary. – Random Variable May 24 '13 at 17:41
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