Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 -

share|improve this question
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

1 Answer 1

up vote 1 down vote accepted

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 $t$.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.