Please may you kindly assist me on this integration exercise: For real $a, b$ with $a \neq 0$, consider the equality $$\int_1^\infty f(x)\sin(a\log \sqrt x)x^b \mathrm{d}x = \int_1^\infty f(x)\sin(a\log \sqrt x)x^{-b} \mathrm{d}x, $$ where $f(x) := \sum_{n=0}^\infty n^{2}\pi (n^{2}\pi - \frac{3}{2x})e^{-n^{2}\pi x + \frac{3}{2}\ln x}$.

Can this equality hold for a nonzero $b$?

  • $\begingroup$ What is $N$? ... $\endgroup$ Jul 26, 2016 at 12:31
  • $\begingroup$ Also, does $3/2x$ mean $3/(2x)$? $\endgroup$ Jul 26, 2016 at 12:32
  • $\begingroup$ $N$ is any real number $>1$, @losif, yes that's it. $\endgroup$
    – QDK
    Jul 26, 2016 at 12:35
  • $\begingroup$ Or we can replace $N$ by $\infty$, which i will do just now in the edit. $\endgroup$
    – QDK
    Jul 26, 2016 at 12:36
  • 1
    $\begingroup$ I mean why didn't you choose e.g. $f(x) = \sum_{n=0}^\infty n\pi (n^{2}\pi - \frac{5}{2x})e^{-n^{3}\pi x + \frac{1}{2}\ln x}$ or $f(x) = \sum_{n=0}^\infty n^{2}\pi (n^{4}\pi - \frac{7}{2x})e^{-n^{2}\pi x + \frac{3}{4}\ln x}$ or anything else? $\endgroup$
    – Stefan Kohl
    Jul 26, 2016 at 13:18

1 Answer 1


Yes. For real $X>1$ and natural $N$, let
\begin{equation} D_{N,X}(a,b):=\int_1^X f_N(x)\sin(a\log \sqrt x)x^b \mathrm{d}x - \int_1^X f_N(x)\sin(a\log \sqrt x)x^{-b} \mathrm{d}x, \end{equation} where \begin{equation} f_N(x):=\sum_{n=0}^{N-1} n^{2}\pi (n^{2}\pi - \frac{3}{2x})e^{-n^{2}\pi x + \frac{3}{2}\ln x}. \end{equation} Note that $D_{N,X}(a,b)$ should be close to $D_{\infty,\infty}(a,b)$ even for moderately large values of $N$ and $X$, since $e^{-n^{2}\pi x}$ decreases exponentially fast in $n$ and $x$.

We compute \begin{equation} D_{15,15}(10,15/2)\approx-4.16514\approx D_{20,20}(10,15/2)<0, \tag{1} \end{equation} \begin{equation} D_{15,15}(10,9)\approx13.7886\approx D_{20,20}(10,9)>0. \tag{2} \end{equation} Now, routine estimates below will show that $D_{\infty,\infty}(10,15/2)<0$ and $D_{\infty,\infty}(10,9)>0$, whence, by continuity, $D_{\infty,\infty}(10,b)=0$ for some $b\in(15/2,9)$. That is, the equality in question holds for $a=10$ and some $b\in(15/2,9)$.

Here are the mentioned routine estimates: for $x\ge1$ and natural $N$, \begin{equation} |f(x)-f_N(x)|\le x^{3/2}\sum_{n=N}^\infty n^{2}\pi (n^{2}\pi)e^{-n^{2}\pi x} \le x^{3/2}\sum_{m=N^2}^\infty (m\pi)^2 e^{-m\pi x} \end{equation} \begin{equation} \le 16x^{-1/2}\sum_{m=N^2}^\infty e^{-m\pi x/2} \le \frac{16}{1-e^{-\pi/2}}\,e^{-N^2\pi x/2}. \end{equation} In particular, $|f(x)|=|f(x)-f_1(x)|\le \frac{16}{1-e^{-\pi/2}}\,e^{-\pi x/2}$ for $x\ge1$, whence \begin{equation} |D_{\infty,X}(a,b)-D_{\infty,\infty}(a,b)|\le \int_X^\infty |f(x)|x^b \mathrm{d}x \le \int_X^\infty \frac{16}{1-e^{-\pi/2}}\,e^{-\pi x/2}x^b \mathrm{d}x <0.21 \end{equation} for $X=20$ and $b\in[0,9]$. It also follows that
\begin{equation} |D_{N,X}(a,b)-D_{\infty,X}(a,b)|\le 2\int_1^\infty |f(x)-f_N(x)|x^b \mathrm{d}x \end{equation} \begin{equation} \le 2\int_0^\infty \frac{16}{1-e^{-\pi/2}}\,e^{-N^2\pi x/2}x^b \mathrm{d}x <10^{-20} \end{equation} for $N=20$ and $b\in[0,9]$. So, for $N=X=20$, all real $a$, and all $b\in[0,9]$ \begin{equation} |D_{N,X}(a,b)-D_{\infty,\infty}(a,b)| \le|D_{N,X}(a,b)-D_{\infty,X}(a,b)|+|D_{\infty,X}(a,b)-D_{\infty,\infty}(a,b)| \end{equation} \begin{equation} <10^{-20}+0.21<0.22<|D_{20,20}(10,15/2)|\wedge|D_{20,20}(10,9)|, \end{equation} by $(1)$--$(2)$. Thus, indeed $D_{\infty,\infty}(10,15/2)<0<D_{\infty,\infty}(10,9)$, whence, by continuity, $D_{\infty,\infty}(10,b)=0$ for some $b\in(15/2,9)$.

  • $\begingroup$ I have added details on the mentioned routine estimates. $\endgroup$ Jul 26, 2016 at 14:49

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