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My question is about what is the problem, if there is one, to get an identity using Frullani's integral. I've in a hand the statement from MathWorld, and other statement from this site, with a nice proof, and a remark in such answer from other user.

Question. It is possible or feasible an application of Frullani's theorem (that is MathWorld) or see here in this site Giraudo's answer for Proof of Frullani's theorem, for the function $$f(x)=\sum_{n=1}^\infty \frac{e^{-\pi n^2 x}}{n^3}?$$

I know that was in the literature the function $g(x)=\sum_{n=1}^\infty e^{-n^2 \pi x}$, since there is a easy relationship with the following evaluation of a Jacobi Theta function $\vartheta_3(0,e^{-\pi x})$, $g(x)=\frac{1}{2}(\vartheta_3(0,e^{-\pi x})-1)$, or that $\lim_{x\to\infty}\vartheta_3(0,e^{-\pi x})=1$. But I know few facts about this kind of functions, I say about convergence questions.

From the first reference, if I take $a=1,b=2$, the symbolic evaluations $f(0)=\zeta(3)$ and $f(\infty)=0$, provide to me

$$\int_0^\infty \frac{1}{x}\left(\sum_{n=1}^\infty \frac{e^{-\pi n^2 x}}{n^3}-\sum_{n=1}^\infty \frac{e^{-2\pi n^2 x}}{n^3}\right)dx=\zeta(3)\log 2.$$

But does converge the integral in LHS? I know that there is a problem at $x=0$ (I've tried several definition for $f(x)$ to get a symbolic identities involving $\zeta(3)$ when one combine easy computations and Frullani's integral, and I don't understand if this current question the theorem provide to me a trivial identity or well I can not use it since the hypothesis of the theorem isn't right).

From the second reference, notice that is requiered that my $f:[0,\infty)\to\mathbb{R}$, and my proble was in my definition for $x=0$, because is required that our functions is continuously differentiable, and the finite value of the limit is not a problem by the remark, that one can redefine the function.

Thanks in advance.

Community
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  • I hope that if it isn't possible to get an application of Frullani theorem, you explain me why, and where were the mistakes. If it is possible an application (maybe my approach was useless) you are welcome to add a variant leaving good details (as your see). –  Jun 22 '16 at 10:43

1 Answers1

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Let we consider: $$ \int_{0}^{+\infty}\color{red}{\frac{1}{x}}\color{blue}{\left(\sum_{n\geq 1}\frac{e^{-\pi n^2 x}-1}{n^3}-\sum_{n\geq 1}\frac{e^{-2\pi n^2 x}-1}{n^3}\right)}\,dx.$$

  1. The integral is convergent by Dirichlet's test, since the red term is decreasing toward zero while the blue term has a bounded primitive: $$\int_{0}^{t}\sum_{n\geq 1}\frac{e^{-\pi n^2 x}-e^{-2\pi n^2 x}}{n^3}\,dx\leq\sum_{n\geq 1}\frac{1}{2\pi n^5}=\frac{\zeta(5)}{2\pi};$$
  2. There is no integrability issue in a right neighbourhood of the origin, since $$\frac{e^{-Cx^2}-1}{x}$$ is an entire function.
Jack D'Aurizio
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  • Very thanks much, now I am trying to understand your claims, very thanks much for your details. –  Jun 22 '16 at 11:54
  • It is appreciated if you can provide to me a description of the statement Frullani's theorem in complex context . Notice that there are very nice answer to this reference, but I would like to know the final statement. that people with experience and knowledges can deduce from your notes and the other users in the question, comments and answer, but it is more difficult for other people. Can you write as a comment here or in the previous post the statement of Frullani Th. in complex context? Thanks . –  Jun 22 '16 at 12:15
  • @user243301: why do you need the complex version? Here the $a$ and $b$ parameters are just positive real numbers. By the way, the statement of the complex version is simple: Frullani's theorem still applies to $$\int_{0}^{+\infty}\frac{e^{-ax}-e^{-bx}}{x},dx $$ provided that $a$ and $b$ are complex numbers with positive real part. – Jack D'Aurizio Jun 22 '16 at 12:31
  • Very thanks much, I only wanted see the final statement. This week I try read the answers of the post to understand what is this complex version. Very thanks much, your help is appreciated. –  Jun 22 '16 at 13:18