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I encountered an astonishing integral (numerically verified)

$$\int_0^{\infty } \frac{1}{\sqrt{x}}\left[\frac{\cos \left(\pi x^2\right)}{\sinh (\pi x)}-\frac{1}{\pi x}\right] \, dx=\frac{1}{\sqrt{2}}\zeta\left(\frac{1}{2}\right)$$

What technique should we use to establish it? Any help will be appreciated.

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  • $\begingroup$ $x^{2}$ is the “bad guy”. $\endgroup$ – Felix Marin Sep 9 '20 at 15:04
  • $\begingroup$ My first thought is something like this; maybe one could write cosine as a series and integrate term-by-term, but that doesn't seem super promising. Convergence is always an issue with these representations as well. $\endgroup$ – FearfulSymmetry Sep 9 '20 at 20:47
  • $\begingroup$ I am thinking that this is a consequence of Ramanujan's master theorem. One just needs to compute the series expansion of $\frac{\cos(\pi x^2)}{\sinh(\pi x)}$ around $x=0$ and then we're done. $\endgroup$ – Shobhit Bhatnagar Sep 14 '20 at 6:58
  • $\begingroup$ @Integrand The linked identity only works when $\operatorname{Re}(s)>1.$ So it wouldn't be of any use in this case. $\endgroup$ – K.defaoite Sep 14 '20 at 21:45
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Let $$\mathscr{I}(s) = \int_0^\infty {{x^{s - 1}}\left( {\frac{{\cos (\pi {x^2})}}{{\sinh \pi x}} - \frac{1}{{\pi x}}} \right)dx} $$ It suffices to prove, for $0<\Re(s)<1$, $$\tag{*}\mathscr{I}(s) + 2\frac{{\Gamma (s)}}{{{{(2\pi )}^s}}}\sin \frac{{\pi s}}{2}\mathscr{I}(1 - s) = 2\frac{{\Gamma (s)\zeta (s)}}{{{{(2\pi )}^s}}}$$


We need a nontrivial Fourier transform:

(Lemma 1) For $\xi\in \mathbb{R}\setminus \{0\}$, $$\int_{ - \infty }^\infty {(\frac{{{e^{i\pi {x^2}}}}}{{\sinh \pi x}} - \frac{1}{{\pi x}}){e^{ - 2\pi ix\xi }}dx} = i\frac{{{e^{ - i\pi {\xi ^2}}} - {e^{\pi \xi }}}}{{\sinh \pi \xi }} + 2i{\chi _{(0,\infty )}}(\xi )$$ where $\chi_A$ is the characteristic function of set $A$.

Proof: Let $C(r)$ be a contour along real axis, but with small indention above $r \in \mathbb{R}$, then $$\int_{C(0)} {\frac{{{e^{i\pi {z^2}}}{e^{ - 2\pi iz\xi }}}}{{\sinh \pi z}}dz} = {e^{ - i\pi {\xi ^2}}}\int_{C( - \xi )} {\frac{{{e^{i\pi {z^2}}}}}{{\sinh \pi (z + \xi )}}dz} $$ Let $F(z)=\dfrac{{{e^{i\pi {z^2}}}{e^{4\pi z}}}}{{\sinh \pi (z + \xi )\sinh 4\pi z}}$, then $$F(z) - F(z + 4i) = \frac{{2{e^{i\pi {z^2}}}}}{{\sinh \pi (z + \xi )}}$$ therefore $$\int_{C( - \xi )} {\frac{{2{e^{i\pi {z^2}}}}}{{\sinh \pi (z + \xi )}}dz} = \int_C {F(z)dz} = 2\pi i\color{red}{\frac{{1 - {e^{\pi \xi }}{e^{i\pi {\xi ^2}}}}}{{\pi \sinh \pi \xi }}}$$ where $C$ is the rectangular contour with vertices $\pm \infty, \pm \infty + 4i$, and has small indentions above $-\xi, -\xi+4i, 0, 4i$. Observe that $F$ has $20$ poles inside $C$, summing over residues at these points gives the red expression. Thus, $$\begin{aligned}&\int_{ - \infty }^\infty {(\frac{{{e^{i\pi {x^2}}}}}{{\sinh \pi x}} - \frac{1}{{\pi x}}){e^{ - 2\pi ix\xi }}dx} = \int_{C(0)} {\frac{{{e^{i\pi {z^2}}}{e^{ - 2\pi iz\xi }}}}{{\sinh \pi z}}dz} - \int_{C(0)} {\frac{{{e^{ - 2\pi iz\xi }}}}{{\pi z}}dz} \\ &= i\frac{{{e^{ - i\pi {\xi ^2}}} - {e^{\pi \xi }}}}{{\sinh \pi \xi }} + 2i{\chi _{(0,\infty )}}(\xi )\end{aligned}$$ proving the lemma.


Let $0<\Re(s)<1$, we have $$\int_0^\infty {{x^{s - 1}}{e^{ - 2\pi ix\xi }}dx} = \frac{{\Gamma (s)}}{{{{(2\pi i)}^s}}}{\xi ^{ - s}}\qquad \Im(\xi)\leq 0$$ Plancherel theorem $\int_\mathbb{R} f(x) \overline{g(x)} dx = \int_\mathbb{R} \hat{f}(\xi) \overline{\hat{g}(\xi)} d\xi$ produces $$\begin{aligned}&\int_0^\infty {{x^{s - 1}}\left( {\frac{{{e^{-i\pi {x^2}}}}}{{\sinh \pi x}} - \frac{1}{{\pi x}}} \right)dx} = \frac{{\Gamma (s)}}{{{{(2\pi i)}^s}}}\int_{ - \infty }^\infty {{\xi ^{ - s}}\left[ { - i\frac{{{e^{i\pi {\xi ^2}}} - {e^{\pi \xi }}}}{{\sinh \pi \xi }} - 2i{\chi _{(0,\infty )}}(\xi )} \right]d\xi } \\ &= - \frac{{\Gamma (s)}}{{{{(2\pi )}^s}}}{e^{ - \pi is/2}}i\int_0^\infty {{\xi ^{ - s}}\left[ {\frac{{{e^{i\pi {\xi ^2}}} - {e^{\pi \xi }}}}{{\sinh \pi \xi }} + 2} \right]d\xi } + \frac{{\Gamma (s)}}{{{{(2\pi )}^s}}}{e^{\pi is/2}}i\int_0^\infty {{\xi ^{ - s}}\frac{{{e^{i\pi {\xi ^2}}} - {e^{ - \pi \xi }}}}{{\sinh \pi \xi }}d\xi }\end{aligned}$$ Taking complex conjugation (i.e. replace $i$ by $-i$), then sum up with the original gives: $$\begin{aligned}\mathscr{I}(s) &= - \frac{{\Gamma (s)}}{{{{(2\pi )}^s}}}\sin \frac{{\pi s}}{2}\int_0^\infty {{\xi ^{ - s}}\left[ {\frac{{{e^{ - i\pi {\xi ^2}}} - {e^{\pi \xi }}}}{{\sinh \pi \xi }} + 2 + \frac{{{e^{i\pi {\xi ^2}}} - {e^{ - \pi \xi }}}}{{\sinh \pi \xi }}} \right]d\xi } \\ & = - \frac{{\Gamma (s)}}{{{{(2\pi )}^s}}}\sin \frac{{\pi s}}{2} \left[ 2\mathscr{I}(1 - s) + 2\int_0^\infty {{x^{ - s}}\left( {\frac{1}{{\pi x}} - \coth \pi x + 1} \right)dx}\right] \end{aligned}$$

the next lemma completes the proof of $(*)$.

(Lemma 2) For $0<\Re(s)<1$, $$\int_0^\infty {{x^{s - 1}}\left( {\frac{1}{{\pi x}} + 1 - \coth \pi x} \right)dx} = - \sec \frac{{\pi s}}{2}\zeta (1 - s)$$

Proof: A lucid method is via Mellin inversion. First, expand $\coth \pi x$ in power series of $e^{-\pi x}$, termwise integration yields $$\int_0^\infty {{x^{s - 1}}\left( {1 - \coth \pi x} \right)dx} = - {2^{1 - s}}{\pi ^{ - s}}\Gamma (s)\zeta (s) = - \sec \frac{{\pi s}}{2}\zeta (1 - s) \qquad \Re(s)>1$$ where we used the functional equation of $\zeta$. Note that $\zeta$ has "moderate growth" in every vertical strip, so Mellin inversion is permissible $$\frac{1}{{2\pi i}}\int_{\sigma - i\infty }^{\sigma + i\infty } {\left[ { - \sec \frac{{\pi s}}{2}\zeta (1 - s)} \right]{x^{ - s}}ds} = 1 - \coth \pi x \qquad \sigma>1$$ Now shift the path of integration, to make it has real part $0<\sigma'<1$, taking residue at $s=1$ into account, the LHS of above equation equals $$ \frac{1}{{2\pi i}}\int_{\sigma ' - i\infty }^{\sigma ' + i\infty } {\left[ { - \sec \frac{{\pi s}}{2}\zeta (1 - s)} \right]{x^{ - s}}ds} - \frac{1}{{\pi x}} \qquad 0<\sigma'<1$$ apply Mellin inversion again proves the lemma.

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    $\begingroup$ All I can say is... Wow. Fantastic work. $\endgroup$ – K.defaoite Sep 15 '20 at 9:18
  • $\begingroup$ excuse my ignorance, but how can you a apply Parseval to an integral over only real half-line without obvious symmetry $x<->-x$? $\endgroup$ – asgeige Sep 15 '20 at 9:18
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    $\begingroup$ @asgeige Plancherel is applied to $f(x) = x^{s-1}\chi_{(0,\infty)}(x)$, its Fourier transform is $$\int_{-\infty}^\infty f(x) e^{-2\pi i x\xi} dx = \int_0^\infty {{x^{s - 1}}{e^{ - 2\pi ix\xi }}dx} = \frac{{\Gamma (s)}}{{{{(2\pi i)}^s}}}{\xi ^{ - s}}$$ so it is really an integral over $\mathbb{R}$, but you make $f$ identically zero at negative reals. $\endgroup$ – pisco Sep 15 '20 at 9:21
  • $\begingroup$ @pisco alright, i see. Thx $\endgroup$ – asgeige Sep 15 '20 at 9:27
  • $\begingroup$ @K.defaoite Thank you for the appreciation. :) $\endgroup$ – pisco Sep 15 '20 at 9:32
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You can make it a little simpler I think.

$$ \begin{align} \frac{1}{\sqrt{x}}\left(\frac{\cos(\pi x^2)}{\sinh(\pi x)} - \frac{1}{\pi x}\right) & = \frac{ e^{-\pi x}}{\pi (e^{\pi x} - e^{-\pi x}) x^{3/2}} - \frac{e^{\pi x}}{\pi (e^{\pi x} - e^{-\pi x}) x^{3/2}} \\ &\quad + \frac{e^{-i \pi x^2}}{(e^{\pi x} - e^{-\pi x}) \sqrt{x}} + \frac{e^{i \pi x^2}}{(e^{\pi x} - e^{-\pi x}) \sqrt{x}} \end{align} $$

These 4 integrals looks slightly easier.

$$\frac{ e^{-\pi x}}{\pi (e^{\pi x} - e^{-\pi x}) x^{3/2}}$$

$$- \frac{e^{\pi x}}{\pi (e^{\pi x} - e^{-\pi x}) x^{3/2}}$$

$$\frac{e^{-i \pi x^2}}{(e^{\pi x} - e^{-\pi x}) \sqrt{x}}$$

$$\frac{e^{i \pi x^2}}{(e^{\pi x} - e^{-\pi x}) \sqrt{x}}$$

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    $\begingroup$ These 4 integrals diverge. $\endgroup$ – metamorphy Sep 11 '20 at 9:17

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