1

The following integral came into my way today.

$${\Large \int_{0}^{\infty} }\frac{\sum \limits_{k=1}^{\infty}k\sin(kx)\,e^{-tk^2}}{\sum \limits_{k=1}^{\infty}\cos(kx)\,e^{-tk^2}}dt=\frac{\pi^2({\pi-x})}{8} \quad , \quad 0<x<2\pi$$

I have no idea how to attack it. First thing coming into my mind is Poisson summation formula but I am not used to using it.. So, basically no clue.

Can we deduce a general form for those two sums in there? I have this strange feeling we can , no?

By the way, since I don't have an attack I am not sure the answer is correct.

Tolaso
  • 6,656
  • 1
    Note that: $$\frac d{dx}\cos(kx)=-k\sin(kx)$$ May come in handy, haven't thought about it much. – Simply Beautiful Art Jun 01 '17 at 20:10
  • Looks like a quotient of theta-functions... – Chappers Jun 01 '17 at 20:12
  • @SimplyBeautifulArt Yeah, I had seen that my self.. but at this stage of the problem which is the beginning I don't see how it helps? – Tolaso Jun 01 '17 at 20:13
  • @Chappers I beg your pardon? Thetas? What is related to thetas? – Tolaso Jun 01 '17 at 20:14
  • @Tolaso: $\sum_{k\geq 1} e^{-tk^2}$ is a Theta. – Jack D'Aurizio Jun 01 '17 at 20:15
  • Ho ho ho , well well well , very nice!! No way I am solving it!! Good to know! – Tolaso Jun 01 '17 at 20:17
  • 1
    Mind I ask where you found the solution? – Simply Beautiful Art Jun 01 '17 at 20:20
  • 1
    The integrand function is of the form $-\frac{f_x}{f}$, hence by integrating with respect to $x$ we get that the integral depends on the logarithm of a theta-like series. My bet is so on a consequence of the Jacobi triple product (https://en.wikipedia.org/wiki/Jacobi_triple_product) – Jack D'Aurizio Jun 01 '17 at 20:20
  • You mean the closed form? Well the problem is stated like this in the leaflet our professor hand out to us today. He said there will be challenging problems but I did not expect them to be that tough!! – Tolaso Jun 01 '17 at 20:21
  • However, in order to bring $$\prod_{m=1}^\infty\left( 1 - x^{2m}\right)\left( 1 + x^{2m-1} y^2\right)\left( 1 +\frac{x^{2m-1}}{y^2}\right)= \sum_{n=-\infty}^\infty x^{n^2} y^{2n}\tag{1}$$ to its extreme consequences, I would be happier to have $1+2\sum_{k\geq 1}\cos(kx)e^{-tk^2}$ in the denominator. – Jack D'Aurizio Jun 01 '17 at 20:28
  • Jack , give the version you mention a try !!! You never know what you shall find !! – Tolaso Jun 01 '17 at 20:34
  • Does not seem to be true when $x=0$ ... Jack's suggestion that the sums should be $ \sum_{k \in \mathbb{Z}}$ might work ? – Donald Splutterwit Jun 01 '17 at 21:32

1 Answers1

3

Houston, we have a problem. The given integral is not well defined: for many fixed values of $x\in(0,2\pi)$, the function

$$ g(t)=\sum_{k\geq 1}\cos(kx)e^{-tk^2} $$ has a simple zero in a right neighbourhood of the origin that is not shared by $\sum_{k\geq 1}k\sin(kx)e^{-tk^2}$.
Simple poles are not integrable singularities, hence the LHS is not well-defined for any $x$ in the range $(0,2\pi)$.

Jack D'Aurizio
  • 353,855
  • How about if we go with the version you propose ?? – Tolaso Jun 01 '17 at 20:55
  • 1
    @Tolaso: before starting a massive amount of probably pointless computations, I would prefer to receive some clarification about the original problem. Can you show to your professor my remark? – Jack D'Aurizio Jun 01 '17 at 20:58
  • Sure !!! However tomorrow !!! It's past midnight here !! – Tolaso Jun 01 '17 at 21:01
  • @Tolaso: no problem, there is no hurry. We will keep in touch. – Jack D'Aurizio Jun 01 '17 at 21:02
  • Hey Jack ... the professor says that he may mistyped the problem and the version you mention is probably the correct one ... he has no further comments this time !! – Tolaso Jun 06 '17 at 19:19