2
$\begingroup$

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 -

  • 1
    @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.2012-04-07

1 Answers 1

1

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 $t$.