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.

Prove that $$\int_0^1 \psi{(x) \sin(2 n \pi x)} \space\mathrm{dx}=-\frac{\pi}{2}, \space n\ge1$$ where $\psi(x)$ - digamma function

share|improve this question
3  
What is $\psi(x)$? –  Qiaochu Yuan Dec 28 '12 at 9:44
1  
The claim is clearly wrong for $n=0$. –  Hagen von Eitzen Dec 28 '12 at 9:53
2  
An interesting commentary for the Riemann-Lebesgue lemma! –  GEdgar Dec 28 '12 at 14:13
add comment

2 Answers

up vote 9 down vote accepted

By the log-differentiation of the Euler's reflection formula, we have

$$ \psi_0(x) - \psi_0(1-x) = -\pi \cot (\pi x). $$

Thus we have

\begin{align*} \int_{0}^{1}\psi_0(x) \sin (2\pi n x) \, dx &= \frac{1}{2}\int_{0}^{1}\psi_0(x) \sin (2\pi n x) \, dx - \frac{1}{2}\int_{0}^{1}\psi_0(1-x) \sin (2\pi n x) \, dx \\ &= -\frac{\pi}{2} \int_{0}^{1} \frac{\sin (2\pi n x)}{\sin (\pi x)} \, \cos (\pi x) \, dx \\ &= -\frac{1}{2} \int_{0}^{\pi} \frac{\sin (2 n \theta)}{\sin \theta} \, \cos \theta \, d\theta \\ &= - \int_{0}^{\frac{\pi}{2}} \frac{\sin (2 n \theta)}{\sin \theta} \, \cos \theta \, d\theta. \end{align*}

Now the rest follows by my blog posting.

share|improve this answer
    
@Chris'ssister, thank you. :) –  sos440 Dec 28 '12 at 10:04
    
@sos440 By the way, cool blog you have out there. –  Sasha Dec 28 '12 at 14:36
    
@sos440: I like your use of the reflection formula. (+1) –  user26872 Dec 29 '12 at 20:08
add comment

Here's another approach using the integral representation for $\psi$. We assume $n$ is an integer greater than or equal to one. Then $$\begin{eqnarray*} \int_0^1 dx\, \sin(2n\pi x) \psi(x) &=& \int_0^1 dx\, \sin(2n\pi x) \int_0^\infty dt\, \left( \frac{e^{-t}}{t} - \frac{e^{-x t}}{1-e^{-t}} \right) \\ &=& \int_0^\infty dt\, \left( \frac{e^{-t}}{t} \int_0^1 dx\, \sin(2n\pi x) - \frac{1}{1-e^{-t}} \int_0^1 dx\, \sin(2n\pi x)e^{-x t} \right). \end{eqnarray*}$$ But $\int_0^1 dx\, \sin(2n\pi x) = 0$ and $$\int_0^1 dx\, \sin(2n\pi x)e^{-x t} = \frac{2n\pi}{t^2+4n^2\pi^2}(1-e^{-t}).$$ (Details for the second integral can be given if necessary.) Therefore $$\begin{eqnarray*} \int_0^1 dx\, \sin(2n\pi x) \psi(x) &=& -\int_0^\infty dt\, \frac{2n\pi}{t^2+4n^2\pi^2} \\ &=& -\frac{\pi}{2}. \end{eqnarray*}$$

share|improve this answer
    
${(\text{precious})}^{\text{precious}}$ (+1) :-) Your way is very short and easy. Thanks! –  Chris's sis Dec 29 '12 at 20:11
    
@Chris'ssister: Glad to help. I had not seen this interesting integral before. (+1) –  user26872 Dec 29 '12 at 21:08
add comment

Your Answer

 
discard

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.