# a gamma integral

any idea how to do this integral ?

$$\lim_{T\rightarrow \infty}\frac{1}{2T}\int_{-T}^{T}\frac{\Gamma(3+it)}{\Gamma(3+it-j)}e^{ikt}dt$$

$j$ is a positive integer. $k$ is a constant - not necessarily positive -

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Where does it comes from and what have you tried? – draks ... Apr 7 '12 at 17:17
Landau's theorem on the coefficients of a Dirichlet series. and i've tried integration by parts . no luck though . – Mohammad Al Jamal Apr 7 '12 at 17:29
With $T\to\infty$, it looks like a Fourier Transform... (this is not meant as a hint, just an observation). You mean the Landau prime ideal theorem? and finally: interesting +1. – draks ... Apr 7 '12 at 18:05
And Landau didn't know how to evaluate it? So you expect that we do? – GEdgar Apr 7 '12 at 18:07
@MohammadAlJamal: perhaps if you find any answers to your previous questions satisfactory, you'd consider selecting one as an "accepted answer" by ticking the checkmark on the left side of the screen. – Tyler Apr 7 '12 at 18:37

For $j=1$ apply the identity $\Gamma(z-1)=\Gamma(z)/(z-1)$ with $z=3+it$ to the denominator to rewrite the problem as $$\frac{1}{2T}\int_{-T}^{T}(2+it)e^{ikt}\,dt = \frac{\cos(kT)}{k} + \frac{(2k-1)\sin(kT)}{k^2T}$$ using integration by parts. This diverges as $T\to\infty$. The expression diverges for larger values of $j$ also --- when $j=2$ for example, the integrand has a quadratic in $t$.