Some definite integrals connected with Gauss's sums
Messenger of Mathematics, XLIV, 1915, 75 – 85

1. If $n$ is real and positive, and $|I(t)|,$ where $I(t)$ is the imaginary part of $t$, is less than either $n$ or $1$, we have

\begin{eqnarray} \int\limits^\infty_0 \frac{\cos \pi tx}{\cosh \pi x} e^{-i \pi n x^2} \ dx &=& 2 \int\limits^\infty_0 \int\limits^\infty_0 \frac{\cos \pi tx \cos 2 \pi x y}{\cosh \pi y} e^{-i \pi n x^2} \ dx dy \nonumber\\ &=& \sqrt{n}~ \exp \left\{- \mbox{$\frac{1}{4}$}i \pi \left(1-\frac{t^2}{n}\right)\right\} \int\limits^\infty_0 \frac{\cos \pi tx}{\cosh \pi n x} e^{i \pi n x^2} \ dx. \end{eqnarray}
When $n=1$ the above formula reduces to
$$\int\limits^\infty_0 \frac{\cos \pi tx}{\cosh \pi x} \sin \pi x^2 \ dx = \tan \{\mbox{\frac{1}{8}}\pi (1-t^2)\} \int\limits^\infty_0 \frac{\cos \pi tx}{\cosh \pi x} \cos \pi x^2 \ dx.$$
if $t=0$, and
\begin{eqnarray} \left.\begin{array}{ll}\ds \phi (n) = \int\limits^\infty_0 \frac{\cos \pi n x^2}{\cosh \pi x} \ dx , \\ \ds\Psi(n) = \int\limits^\infty_0 \frac{\sin \pi n x^2}{\cosh \pi x} \ dx, \end{array}\right\} \end{eqnarray}
then
\begin{equation*} \left.\begin{array}{ll} \phi (n) = \sqrt{ \left(\frac{2}{n}\right) \Psi \left(\frac{1}{n}\right) + \Psi (n)}, \\ \Psi(n) = \sqrt{\left(\frac{2}{n}\right) \phi \left(\frac{1}{n}\right) - \phi(n)}. \end{array}\right\}\tag{3'} \end{equation*}
Similarly, if $\frac{1}{2} \sqrt{3}|I(t)|$ is less than either 1 or $n$, we have
\begin{eqnarray} \int\limits^\infty_0 && \frac{\cos \pi tx}{1+2 \cosh (2 \pi x/\sqrt{3})} e^{i \pi n x^2} \ dx \nonumber\\ &=& \sqrt{n} ~ \exp \left\{-\mbox{$\frac{1}{4}$} i \pi \left(1- \frac{t^2}{n}\right)\right\} \int\limits^\infty_0 \frac{\cos \pi tx}{1+2 \cosh (2 \pi n x/ \sqrt{3})} e^{- i \pi n x^2} \ dx. \end{eqnarray}
If in (4) we suppose $n=1$, we obtain
$$\int\limits^\infty_0 \frac{\cos \pi tx \sin \pi x^2}{1+2 \cosh (2 \pi x/\sqrt{3})} \ dx =\tan\{\mbox{\frac{1}{8}}\pi (1- t^2)\}\int\limits^\infty_0\frac{\cos\pi tx \cos \pi x^2}{1+2 \cosh (2 \pi x/ \sqrt{3})} \ dx;$$
and if $t=0$, and
\begin{eqnarray} \left.\begin{array}{ll} \ds\phi (n) = \int\limits^\infty_0 \frac{\cos \pi n x^2}{1+2 \cosh (2 \pi x/ \sqrt{3})} \ dx, \\ \ds\Psi (n) = \int\limits^\infty_0 \frac{\sin \pi n x^2}{1+2 \cosh (2 \pi x/\sqrt{3})} \ dx, \end{array}\right\} \end{eqnarray}
then
\begin{equation*} \left.\begin{array}{ll} \phi(n) = \sqrt{\left(\frac{2}{n}\right) \Psi \left(\frac{1}{n}\right) + \Psi (n)}, \\ \Psi (n) = \sqrt{\left(\frac{2}{n}\right) \phi \left(\frac{1}{n}\right) - \phi(n)}. \end{array}\right\}\tag{6'} \end{equation*}
In a similar manner we can prove that
$$\int\limits^\infty_0 \frac{\sin \pi tx}{\tanh \pi x} e^{- i\pi n x^2} \ dx = - \sqrt{n} ~ \exp \left\{\mbox{\frac{1}{4}} i \pi \left(1+\frac{t^2}{n}\right)\right\} \int\limits^\infty_0 \frac{\sin \pi tx}{\tanh \pi n x} e^{i \pi n x^2} \ dx.$$
If we put $n=1$ in (7), we obtain
$$\int\limits^\infty_0 \frac{\sin \pi tx}{\tanh \pi x} \cos \pi x^2 \ dx = \tan \{\frac{1}{8} \pi (1+t^2)\} \int\limits^\infty_0 \frac{\sin \pi tx}{\tanh \pi x} \sin \pi x^2 \ dx.$$
Now
\begin{eqnarray} \lim_{t \to 0} \frac{1}{t} \int\limits^\infty_0 \frac{\sin a tx}{\tanh bx} e^{i c x^2} \ dx &=& \lim_{t \to 0} \frac{1}{t} \int\limits^\infty_0 \frac{2 \sin atx}{e^{2bx} - 1} e^{i c x^2} \ dx + \lim_{t \to 0} \int\limits^\infty_0 \frac{\sin a t x}{t} e^{i c x^2} \ dx \nonumber \\ &=& \int\limits^\infty_0 \frac{ae^{icx}}{e^{2b\sqrt{x}} - 1} \ dx + \frac{ia}{2c}. \end{eqnarray}
Hence, dividing both sides of (7) by $t$, and making $t \to 0$, we obtain the result corresponding to (3) and (6), viz.: if
\begin{eqnarray} \left.\begin{array}{ll}\ds \phi(n) = \int\limits^\infty_0 \frac{\cos \pi n x}{e^{2 \pi \sqrt{x}}-1} \ dx, \\ \ds\Psi (n) = \frac{1}{2 \pi n} + \int\limits^\infty_0 \frac{\sin \pi nx}{e^{2 \pi \sqrt{x}}-1} \ dx, \end{array}\right\} \end{eqnarray}
then
\begin{equation*} \left.\begin{array}{ll} \phi(n) = \frac{1}{n} \sqrt{\left(\frac{2}{n}\right) \Psi \left(\frac{1}{n} \right) - \Psi(n)}, \\ \Psi(n) = \frac{1}{n} \sqrt{\left(\frac{2}{n}\right) \phi \left(\frac{1}{n}\right) + \phi(n)}. \end{array}\right\} \tag{10'} \end{equation*}

2. I shall now shew that the integral (1) may be expressed in finite terms for all rational values of $n$. Consider the integral $$J(t) = \int\limits^\infty_0 \frac{\cos tx}{\cosh \frac{1}{2} \pi x} \frac{ \ dx}{a^2 + x^2}.$$ If $R(a)$ and $t$ are positive, we have

\begin{eqnarray} J(t) &=& \frac{4}{\pi} \int\limits^\infty_0 \sum^{r=\infty}_{r=0} \frac{(-1)^r (2r+1)}{x^2 + (2r+1)^2} \frac{\cos tx}{a^2 + x^2} \ dx \nonumber\\ &=& 2 \sum^{r=\infty}_{r=0} \frac{(-1)^r}{a^2-(2r+1)^2} \left\{e^{-(2r+1)t} - \frac{1}{a} (2r+1) e^{-at}\right\} \nonumber \\ &=& \frac{\pi e^{-at}}{2a \cos \frac{1}{2} \pi a} + 2 \sum^{r=\infty}_{r=0} \frac{(-1)^r e^{-(2r+1)t}}{a^2 - (2r+1)^2}, \end{eqnarray}
and it is easy to see that this last equation remains true when $t$ is complex, provided $R(t)>0$ and $|I(t)|\leq \frac{1}{2} \pi$. Thus the integral $J(t)$ can be expressed in finite terms for all rational values of $a$. Thus, for example, we have
\begin{eqnarray} \left.\begin{array}{ll} \ds\int\limits^\infty_0 \frac{\cos tx}{\cosh \frac{1}{2} \pi x} \frac{ \ dx}{1+x^2} = \cosh t \log (2 \cosh t) - t \sinh t, \\ \ds\int\limits^\infty_0 \frac{\cos 2 tx}{\cosh \pi x} \frac{ \ dx}{1+x^2} = 2 \cosh t - (e^{2t} \tan^{-1} e^{-t} + e^{-2t} \tan^{-1} e^t), \end{array}\right\} \end{eqnarray}
and so on. Now let
$$F(n) = \int\limits^\infty_0 \frac{\cos 2 tx}{\cosh \pi x} e^{- i \pi n x^2} \ dx.$$
Then, if $R (a) > 0,$
$$\int\limits^\infty_0 e^{-an} F(n) dn = \int\limits^\infty_0 \frac{\cos 2 tx}{\cosh \pi x} \frac{ \ dx}{a + i \pi x^2} .$$
Now let
\begin{eqnarray} f(n) &=& \sum^{r=\infty}_{r=0} (-1)^r \exp \{-(2r+1) t + \mbox{$\frac{1}{4}$} (2r +1)^2 i \pi n\} \nonumber\\ &+& \frac{1}{\sqrt{n}} \exp \left\{- i \left(\mbox{$\frac{1}{4}$} \pi - \frac{t^2}{\pi n}\right)\right\} \sum^{r=\infty}_{r=0} (-1)^r \exp \left\{- (2r+1) \frac{t}{n} \right.%\ncr \left.- \mbox{$\frac{1}{4}$} (2r+1)^2 \frac{i \pi}{n}\right\}. \end{eqnarray}
Then
\begin{eqnarray} \int\limits^\infty_0 e^{-an} f(n) dn &=& \sum^{r=\infty}_{r=0} \frac{(-1)^r e^{-(2r+1)t}}{a-\mbox{$\frac{1}{4}$} (2r+1)^2 i \pi} + \sqrt{\left(\frac{\pi}{2a}\right)} \frac{\exp \{-\sqrt{(2a/\pi)}(1-i)t\}}{(1+i) \cosh \{(1+i)\sqrt{(\frac{1}{2} \pi a)}\}} \nonumber\\ &=& \int\limits^\infty_0 \frac{\cos 2 tx}{\cosh \pi x} \frac{ \ dx}{a+i\pi x^2}, \end{eqnarray}
in virtue of (11); and therefore
$$\int\limits^\infty_0 e^{-an} \{F(n) - f(n) \} dn = 0.$$
Now it is known that, if $\phi (n)$ is continuous and $$\int\limits^\infty_0 e^{-an} \phi (n) dn = 0,$$ for all positive values of $a$ (or even only for an infinity of such values in arithmetical progression), then $$\phi (n) = 0,$$ for all positive values of $n$. Hence
$$F(n) = f(n).$$
Equating the real and imaginary parts in (13) and (15) we have
\begin{eqnarray} \int\limits^\infty_0 && \frac{\cos 2tx}{\cosh \pi x} \cos \pi n x^2 \ dx = \left\{e^{-t} \cos \frac{\pi n}{4} - e^{-3t} \cos \frac{9 \pi n}{4} + e^{-5t} \cos \frac{25 \pi n}{4} - \cdots \right\} \nonumber\\ &+& \frac{1}{\sqrt{n}} \left\{e^{-t/n} \cos \left(\frac{\pi}{4} - \frac{t^2}{\pi n} + \frac{\pi}{4n}\right) - e^{-3t/n} \cos \left(\frac{\pi}{4} - \frac{t^2}{\pi n} + \frac{9 \pi}{4n} \right) + \cdots\right\}, \end{eqnarray}
\begin{eqnarray} \int\limits^\infty_0 && \frac{\cos 2 tx}{\cosh \pi x} \sin \pi n x^2 \ dx = - \left\{e^{-t} \sin \frac{\pi n}{4} - e^{-3t} \sin \frac{9 \pi n}{4} + e^{-5t} \sin \frac{25 \pi n}{4} - \cdots \right\} \nonumber \\ &+& \frac{1}{\sqrt{n}} \left\{e^{-t/n} \sin \left(\frac{\pi}{4} - \frac{t^2}{\pi n} + \frac{\pi}{4n} \right) - e^{-3t/n} \sin \left(\frac{\pi}{4} - \frac{t^2}{\pi n} + \frac{9 \pi}{4n} \right) + \cdots \right\}. \end{eqnarray}
We can verify the results (18), (19) and (20) by means of the equation (1). This equation can be expressed as a functional equation in $F(n)$, nd it is easy to see that $f(n)$ satisfies the same equation. The right-hand side of these equations can be expressed in finite terms if $n$ is any rational number. For let $n=a/b,$ where $a$ and $b$ are any two positive integers and one of them is odd. Then the results (19) and (20) reduce to
\begin{eqnarray} 2 \cosh bt && \int\limits^\infty_0 \frac{\cos 2 tx}{\cosh \pi x} \cos \left(\frac{\pi a x^2}{b}\right) \ dx \nonumber\\ &=& [ \cosh \{(1-b)t\} \cos (\pi a/4b) - \cosh \{(3-b)t\} \cos (9 \pi a/4b) \nonumber\\ &+& \cosh \{(5-b)t\} \cos (25 \pi a/4b) - \cdots \mbox{ to } b \mbox{ terms }]\nonumber \\ &+& \sqrt{\left(\frac{b}{a}\right)} \left[\cosh \left\{\left(1-\frac{1}{a}\right)bt\right\} \cos \left(\frac{\pi}{4} - \frac{bt^2}{\pi a} + \frac{\pi b}{4a}\right)\right. \nonumber\\ &-& \left. \cosh \left\{\left(1-\frac{3}{a}\right) bt \right\} \cos \left(\frac{\pi}{4} - \frac{bt^2}{\pi a} + \frac{9 \pi b}{4a}\right) + \cdots \mbox{ to } a \mbox{ terms }\right], \end{eqnarray}
\begin{eqnarray} 2 \cosh bt && \int\limits^\infty_0 \frac{\cos 2 tx}{\cosh \pi x} \sin \left(\frac{\pi a x^2}{b}\right) \ dx \nonumber\\ &=& - [ \cosh \{(1-b)t\} \sin (\pi a/4b) - \cosh \{(3-b)t\} \sin (9 \pi a/4b)\nonumber \\ &+& \cosh \{(5-b) t\} \sin (25 \pi a/4b) - \cdots \mbox{ to } b \mbox{ terms }] \nonumber\\ &+& \sqrt{\left(\frac{b}{a}\right)} \left[ \cosh \left\{\left(1-\frac{1}{a}\right)bt\right\} \sin \left(\frac{\pi}{4}-\frac{bt^2}{\pi a} + \frac{\pi b}{4a}\right)\right.\nonumber \\ &-& \left. \cosh \left\{\left(1-\frac{3}{a}\right)bt\right\} \sin \left(\frac{\pi}{4} - \frac{bt^2}{\pi a} + \frac{9 \pi b}{4a}\right) + \cdots \mbox{ to } a \mbox{ terms } \right]. \end{eqnarray}

Thus, for example, we have, when $a=1$ and $b=1$,

$$\int\limits^\infty_0 \frac{\cos \pi x^2}{\cosh \pi x} \cos 2 \pi tx \ dx =\frac{1+\sqrt{2} \sin \pi t^2}{2 \sqrt{2} \cosh \pi t},$$
$$\int\limits^\infty_0 \frac{\sin \pi x^2}{\cosh \pi x} \cos 2 \pi tx \ dx = \frac{-1 + \sqrt{2} \cos \pi t^2}{2 \sqrt{2} \cosh \pi t}.$$
It is easy to verify that (23) and (24) satisfy the relation (2). The values of the integrals $$\int\limits^\infty_0 \frac{\cos \pi n x^2}{\cosh \pi x} \ dx, \int\limits^\infty_0 \frac{\sin \pi n x^2}{\cosh \pi x} \ dx$$ can be obtained easily from the preceding results by putting $t=0$, and need no special discussion. By successive differentiations of the results (19) and (20) with respect to $t$ and $n$, we can evaluate the integrals
\begin{eqnarray} \left.\begin{array}{ll} \ds\int\limits^\infty_0 x^{2m-1} \frac{\sin tx}{\cosh \pi x} \begin{array}{ll}\cos\\ \sin\end{array} \pi n x^2 \ dx,\\ \ds\int\limits^\infty_0 x^{2m} \frac{\cos tx}{\cosh \pi x} \begin{array}{ll} \cos \\ \sin \end{array} \pi n x^2 \ dx, \end{array}\right\} \end{eqnarray}
for all rational values of $n$ and all positive integral values of $m$. Thus, for example, we have
\begin{eqnarray} \left.\begin{array}{ll} \ds\int\limits^\infty_0 x^2 \frac{\cos \pi x^2}{\cosh \pi x} \ dx = \frac{1}{8\sqrt{2}} - \frac{1}{4 \pi}, \\ \ds\int\limits^\infty_0 x^2 \frac{\sin \pi x^2}{\cosh \pi x} \ dx = \frac{1}{8} - \frac{1}{8 \sqrt{2}}. \end{array}\right\} \end{eqnarray}

3. We can get many interesting results by applying the theory of Cauchy's reciprocal functions to the preceding results. It is known that, if

$$\int\limits^\infty_0 \phi(x) \cos kn x \ dx = \Psi (n),$$
then
\begin{eqnarray*} \mbox{(i) }\frac{1}{2} \alpha && \left\{\frac{1}{2} \phi (0) + \phi (\alpha) + \phi (2 \alpha) + \phi (3 \alpha) + \cdots \right\}\\ &=& \frac{1}{2} \Psi (0) + \Psi (\beta) + \Psi (2 \beta) + \Psi (3 \beta) + \cdots,\tag{27} \end{eqnarray*}
with the condition $\alpha \beta = 2 \pi/k$;
\begin{eqnarray*} \mbox{(ii) }\alpha \sqrt{2} \{\phi (\alpha) - \phi (3 \alpha) &-& \phi (5 \alpha) + \phi (7 \alpha) + \phi (9 \alpha) - \cdots \} \\ &=& \Psi (\beta) - \Psi(3 \beta) - \Psi (5 \beta) + \Psi (7 \beta) + \Psi (9 \beta)- \cdots,\tag{27} \end{eqnarray*}
with the condition $\alpha \beta = \pi/4k$;
\begin{eqnarray*} \mbox{(iii) }\alpha \sqrt{3} \{\phi (\alpha) - \phi(5 \alpha) &-& \phi(7 \alpha) + \phi (11 \alpha) + \phi (13 \alpha) - \cdots \}\\ &=& \Psi (\beta) - \Psi (5 \beta) - \Psi (7 \beta) + \Psi (11 \beta) + \Psi(13 \beta) - \cdots,\tag{27} \end{eqnarray*}
with the condition $\alpha \beta = \pi/6k,$ where 1, 5, 7, 11, 13, $\ldots$ are the odd natural numbers without the multiples of 3. There are of course corresponding results for the function
$$\int\limits^\infty_0 \phi (x) \sin kn x \ dx = \Psi (n),$$
such as $$\alpha \{ \phi (\alpha) - \phi (3 \alpha) + \phi(5 \alpha) - \cdots \} = \Psi (\beta) - \Psi (3 \beta) + \Psi (5 \beta) - \cdots ,$$ with the condition $\alpha \beta = \pi/2k.$ Thus from (23) and (27) (i) we obtain the following results. If

$$F(\alpha, \beta) = \sqrt{\alpha} \left\{\frac{1}{2} + \sum^{r=\infty}_{r=1} \frac{\cos r^2 \pi \alpha^2}{\cosh r \pi \alpha}\right\} - \sqrt{\beta} \sum^{r=\infty}_{r=1} \frac{\sin r^2 \pi \beta^2}{\cosh r \pi \beta},$$
then $$F(\alpha, \beta) = F(\beta, \alpha) = \sqrt{(2\alpha)} \{\frac{1}{2} + e^{-\pi \alpha} + e^{-4\pi\alpha} + e^{-9 \pi \alpha} + \cdots \}^2,$$ provided that $\alpha \beta = 1$.

4. If, instead of starting with the integral (11), we start with the corresponding sine integral, we can shew that, when $R(a)$ and $R(t)$ are positive and $|I(t)| \leq \pi$,

$$\int\limits^\infty_0 \frac{\sin tx}{\sinh \pi x} \frac{ \ dx}{a^2+x^2} = \frac{1}{2a^2} - \frac{\pi e^{-at}}{2 a \sin \pi a} + \sum^{r=\infty}_{r=1} \frac{(-1)^r e^{-rt}}{a^2-r^2}.$$
Hence the above integral can be expressed in finite terms for all rational values of $a$. For example, we have
$$\int\limits^\infty_0 \frac{\sin tx}{\sinh \frac{1}{2} \pi x} \frac{ \ dx}{1+x^2} = e^t \tan^{-1} e^{-t} - e^{-t} \tan^{-1} e^t.$$
From (30) we can deduce that
\begin{eqnarray} \int\limits^\infty_0 \frac{\sin 2tx}{\sinh \pi x} e^{- i \pi n x^2} \ dx &=& \frac{1}{2} - e^{-2t + i\pi n} + e^{-4t+4i\pi n}- e^{-6t + 9 i \pi n} +\cdots \nonumber\\ &-& \frac{1}{\sqrt{n}} \exp \left\{\left(\mbox{$\frac{1}{4}$} \pi + \frac{t^2}{\pi n}\right)i \right\} \{e^{-(t+\frac{1}{4} i \pi)/n} + e^{-(3t + \frac{9}{4} i \pi)/n} + \cdots \}, \end{eqnarray}
$R(t)$ being positive and $|I(t)|\leq \frac{1}{2} \pi$. The right-hand side can be expressed in finite terms for all rational values of $n$. Thus, for example, we have
$$\int\limits^\infty_0 \frac{\cos \pi x^2}{\sinh \pi x} \sin 2 \pi tx \ dx = \frac{\cosh \pi t-\cos \pi t^2}{2 \sinh \pi t},$$
$$\int\limits^\infty_0 \frac{\sin \pi x^2}{\sinh \pi x} \sin 2 \pi tx \ dx = \frac{\sin \pi t^2}{2 \sinh \pi t},$$
and so on. Applying the formula (28) to (33) and (34), we have, when $\alpha \beta = \frac{1}{4}$,
\begin{eqnarray} \left.\begin{array}{lll} \displaystyle{ \sqrt{\alpha} \sum^{r=\infty}_{r=0} (-1)^r \frac{\cos\{(2r +1)^2\pi \alpha^2\}}{\sinh \{(2r+1)\pi \alpha\}}} &+& \displaystyle{\sqrt{\beta} \sum^{r=\infty}_{r=0} (-1)^r \frac{\cos \{(2r+1)^2 \pi \beta^2\}}{\sinh \{(2r+1)\pi \beta\}}}\\ &=& \displaystyle{2 \sqrt{\alpha} \{\frac{1}{2} + e^{-2 \pi \alpha} + e^{- 8 \pi \alpha} + e^{-18 \pi \alpha} + \cdots \}^2}; \\ \displaystyle{\sqrt{\alpha} \sum^{r=\infty}_{r=0} (-1)^r \frac{\sin \{(2r+1)^2 \pi \alpha^2\}}{\sinh \{(2r+1)\pi \alpha\}}} &=& \displaystyle{\sqrt{\beta} \sum^{r=\infty}_{r=0} (-1)^r \frac{\sin \{(2r+1)^2 \pi \beta^2\}}{\sinh \{(2r+1)\pi \beta\}}}. \end{array}\right\} \end{eqnarray}
By successive differentiation of (32) with respect to $t$ and $n$ we can evaluate the integrals
\begin{eqnarray} \left.\begin{array}{lll} \displaystyle{ \int\limits^\infty_0 x^{2m-1} \frac{\cos tx}{\sinh \pi x} \begin{array}{c} \cos \\ \sin \end{array} \pi n x^2 \ dx,} \\ \displaystyle{\int\limits^\infty_0 x^{2m} \frac{\sin tx}{\sinh \pi x} \begin{array}{c} \cos\\ \sin\end{array} \pi n x^2 \ dx} \end{array}\right\} \end{eqnarray}
for all rational values of $n$ and all positive integral values of $m$. Thus, for example, we have
\begin{eqnarray} \left.\begin{array}{ll}\displaystyle{ \int\limits^\infty_0 x \frac{\cos \pi x^2}{\sinh \pi x} \ dx = \frac{1}{8}}, & \displaystyle{\int\limits^\infty_0 x \frac{\sin \pi x^2}{\sinh \pi x} \ dx = \frac{1}{4 \pi}}, \\ \displaystyle{\int\limits^\infty_0 x^3 \frac{\cos \pi x^2}{\sinh \pi x} \ dx = \frac{1}{16} \left(\mbox{$\frac{1}{4}$} - \frac{3}{\pi^2}\right)}, & \displaystyle{\int\limits^\infty_0 x^3 \frac{\sin \pi x^2}{\sinh \pi x} \ dx = \frac{1}{16 \pi}}, \end{array}\right\} \end{eqnarray}
and so on. The denominators of the integrands in (25) and (36) are $\cosh \pi x$ and $\sinh \pi x$. Similar integrals having the denominators of their integrands equal to $$\prod^r_1 \cosh \pi a_r x \sinh \pi b_r x$$ can be evaluated, if $a_r$ and $b_r$ are rational, by splitting up the integrand into partial fractions.

5. The preceding formulĂ¦ may be generalised. Thus it may be shewn that, if $R(a)$ and $R(t)$ are positive, $|I(t)|\leq \pi$, and $-1 \lt R(\theta) \lt 1$, then

\begin{eqnarray} \sin \pi \theta && \int\limits^\infty_0 \frac{\cos tx}{\cosh \pi x + \cos \pi \theta} \frac{ \ dx}{a^2 + x^2}\nonumber\\ &=& \frac{\pi}{2a} \frac{e^{-at} \sin \pi \theta}{\cos \pi a + \cos \pi \theta} + \sum^{r=\infty}_{r=0} \left\{\frac{e^{-(2r+1-\theta)t}}{a^2 - (2r+1-\theta)^2} - \frac{e^{-(2r+1+\theta)t}}{a^2-(2r+1+\theta)^2}\right\}. \end{eqnarray}
From (38) it can be deduced that, if $n$ and $R(t)$ are positive, $|I(t)|\leq \pi$, and $-1 \lt \theta \lt 1$, then
\begin{eqnarray} \sin \pi \theta && \int\limits^\infty_0 \frac{\cos tx}{\cosh \pi x + \cos \pi \theta} e^{- i \pi n x^2} \ dx \nonumber\\ &=& \sum^{r=\infty}_{r=0} \{e^{-(2r+1-\theta) t+ (2r+1-\theta)2 i \pi n} - e^{-(2r+1+\theta) t+ (2r+1+\theta) 2 i \pi n}\}\nonumber\\ &+& \frac{1}{\sqrt{n}} \exp \left\{-\mbox{$\frac{1}{4}$} i \left(\pi - \frac{t^2}{\pi n}\right)\right\} \sum^{r=\infty}_{r=1} (-1)^{r-1} \sin r \pi \theta e^{- (2r t+r^2 i \pi)/4n}. \end{eqnarray}
The right-hand side can be expressed in finite terms if $n$ and $\theta$ are rational. In particular, when $\theta = \frac{1}{3}$, we have
\begin{eqnarray} \int\limits^\infty_0 && \frac{\cos tx}{1+2 \cosh (2 \pi x/\sqrt{3})} e^{-i \pi n x^2} \ dx \nonumber \\ &=& \frac{1}{2} \{e^{-\frac{1}{3} (t \sqrt{3} - i\pi n)} - e^{-\frac{1}{3}(2t \sqrt{3} - 4i \pi n)} + e^{-\frac{1}{3}(4t \sqrt{3}-16 i \pi n)} - \cdots \}\nonumber \\ &+& \frac{1}{2\sqrt{n}}\exp \left\{- \mbox{$\frac{1}{4}$} i \left(\pi - \frac{t^2}{\pi n}\right)\right\}\nonumber \\ && \{e^{-(t\sqrt{3}+i \pi )/3n} - e^{-(2t\sqrt{3}+ 4 i \pi)/3n} + e^{-(4t \sqrt{3} + 16 i \pi)/3n} - \cdots \}, \end{eqnarray}
where 1,2,4,5, $\ldots$ are the natural numbers without the multiples of 3. As an example, when $n=1$, we have
\begin{eqnarray} \left.\begin{array}{ll} \displaystyle{ \int\limits^\infty_0 \frac{\cos \pi x^2 \cos \pi tx}{1+2 \cosh (2 \pi x/\sqrt{3})} \ dx = \frac{1-2 \sin \{(\pi - 3 \pi t^2)/12\}}{8 \cosh (\pi t/ \sqrt{3})-4},} \\ \displaystyle{\int\limits^\infty_0 \frac{\sin \pi x^2 \cos \pi tx}{1+2 \cosh (2 \pi x/\sqrt{3})} \ dx = \frac{-\sqrt{3} + 2 \cos \{(\pi - 3 \pi t^2)/12\}}{8 \cosh (\pi t/\sqrt{3})-4}}. \end{array}\right\} \end{eqnarray}

6. The formula (32) assumes a neat and elegant form when $t$ is changed to $t+\frac{1}{2} i \pi$. We have then

\begin{eqnarray} \int\limits^\infty_0 && \frac{\sin tx}{\tanh \pi x} e^{-i \pi n x^2} \ dx \quad\quad (n > 0, t > 0) \nonumber\\ &=& \left\{\frac{1}{2} + e^{-t+i\pi n} + e^{-2t+4i\pi n} + e^{-3t+9i \pi n} + \cdots \right\} \nonumber\\ &-& \frac{1}{\sqrt{n}} \exp \left\{\mbox{$\frac{1}{4}$} i \left(\pi + \frac{t^2}{\pi n}\right) \right\} \left\{\frac{1}{2} + e^{-(t+i \pi)/n} + e^{-(2t+4 i \pi)/n} + \cdots \right\}. \end{eqnarray}
In particular, when $n=1$, we have
\begin{eqnarray} \left.\begin{array}{ll} \displaystyle{\int\limits^\infty_0 \frac{\cos \pi x^2}{\tanh \pi x} \sin 2 \pi tx \ dx = \frac{1}{2} \tanh \pi t \{1-\cos (\mbox{$\frac{1}{4}$} \pi + \pi t^2)\}}, \\ \ds{\int\limits^\infty_0 \frac{\sin \pi x^2}{\tanh \pi x} \sin 2 \pi tx \ dx = \frac{1}{2} \tanh \pi t \sin (\mbox{$\frac{1}{4}$} \pi+ \pi t^2).} \end{array}\right\} \end{eqnarray}
We shall now consider an important special case of (42). It can easily be seen from (9) that the left-hand side of (42), when divided by $t$, tends to
$$\int\limits^\infty_0 \frac{\cos \pi n x}{e^{2 \pi \sqrt{x}}-1} \ dx - i \left\{\frac{1}{2 \pi n} + \int\limits^\infty_0 \frac{\sin \pi n x}{e^{2 \pi \sqrt{x}}-1} \ dx \right\}$$
as $t \to 0$. But the limit of the right-hand side of (42) divided by $t$ can be found when $n$ is rational. Let then $n = a/b$, where $a$ and $b$ are any two positive integers, and let $$\phi (n) = \int\limits^\infty_0 \frac{\cos \pi n x}{e^{2 \pi \sqrt{x}}-1} \ dx, \Psi (n) = \frac{1}{2 \pi n} + \int\limits^\infty_0 \frac{\sin \pi n x}{e^{2\pi \sqrt{x}}-1} \ dx.$$ The relation between $\phi(n)$ and $\Psi(n)$ has been stated already in (10'). From (42) and (44) it can easily be deduced that, if $a$ and $b$ are both odd, then
\begin{eqnarray} \left.\begin{array}{ll} \displaystyle{ \phi \left(\frac{a}{b}\right) = \mbox{$\frac{1}{4}$} \sum^{r=b}_{r=1} (b-2r) \cos \left(\frac{r^2 \pi a}{b}\right) - \frac{b}{4a} \sqrt{\left(\frac{b}{a}\right)} \sum^{r=a}_{r=1} (a-2r)\sin \left(\mbox{$\frac{1}{4}$} \pi + \frac{r^2 b \pi}{a}\right),} \\ \displaystyle{ \Psi \left(\frac{a}{b}\right) = - \mbox{$\frac{1}{4}$} \sum^{r=b}_{r=1} (b-2r) \sin \left(\frac{r^2 \pi a}{b}\right) + \frac{b}{4a} \sqrt{\left(\frac{b}{a}\right)} \sum^{r=a}_{r=1} (a-2r) \cos \left(\mbox{$\frac{1}{4}$} \pi + \frac{r^2 \pi b}{a} \right),}\end{array}\right\} \end{eqnarray}
It can easily be seen that these satisfy the relation (10'). Similarly, when one of $a$ and $b$ is odd and the other even, it can be shewn that
\begin{eqnarray} \left.\begin{array}{lll} \phi \left(\frac{a}{b}\right) &=& \frac{\sigma}{4 \pi a \sqrt{a}} - \frac{1}{2} \sum^{r=b}_{r=1} r \left(1-\frac{r}{b}\right) \cos \left(\frac{r^2 \pi a}{b}\right)\\ &+& \frac{b}{2a} \sqrt{\left(\frac{b}{a}\right)} \sum^{r=a}_{r=1} r\left(1-\frac{r}{a}\right) \sin \left(\mbox{$\frac{1}{4}$} \pi + \frac{r^2 \pi b}{a}\right), \\ \Psi \left(\frac{a}{b}\right) &=& \frac{\sigma'}{4 \pi a \sqrt{a}} + \frac{1}{2} \sum^{r=b}_{r=1} r\left(1-\frac{r}{b}\right) \sin \left(\frac{r^2 \pi a}{b}\right) \\ &-& \frac{b}{2a} \sqrt{\left(\frac{b}{a}\right)} \sum^{r=a}_{r=1} r \left(1-\frac{r}{a}\right) \cos \left(\mbox{$\frac{1}{4}$}\pi + \frac{r^2 \pi b}{a}\right),\end{array}\right\} \end{eqnarray}
where
\begin{eqnarray} \left.\begin{array}{ll} \sigma =\sqrt{b}\sum_1^a \cos\left (\mbox{$\frac{1}{4}$}\pi+\frac{r^2\pi b}{a}\right )=\sqrt{a}\sum_1^b \sin\left (\frac{r^2\pi a}{b}\right ),\\ \sigma' =\sqrt{b}\sum_1^a \sin\left (\mbox{$\frac{1}{4}$}\pi+\frac{r^2\pi b}{a}\right )=\sqrt{a}\sum_1^b \cos\left (\frac{r^2\pi a}{b}\right ).\end{array}\right\} \end{eqnarray}
Thus, for example, we have
\begin{eqnarray} \left.\begin{array}{ll} \phi (0) = \frac{1}{12}, \phi (1) = \frac{2-\sqrt{2}}{8}, \phi (2) = \frac{1}{16}, \phi (4) = \frac{3-\sqrt{2}}{32}, \\ \phi(6) = \frac{13-4\sqrt{3}}{144}, \phi\left(\frac{1}{2}\right) = \frac{1}{4 \pi}, \phi \left(\frac{2}{5}\right) = \frac{8-3 \sqrt{5}}{16},\end{array}\right\} \end{eqnarray}
and so on. By differentiating (42) with respect to $n$, we can evaluate the integrals
$$\int\limits^\infty_0 \frac{x^m}{e^{2 \pi \sqrt{x}}-1} \begin{array}{l}\cos \\ \sin \end{array} \pi n x \ dx$$
for all rational values of $n$ and positive integral values of $m$. Thus, for example, we have
\begin{eqnarray} \left.\begin{array}{lll} \ds\int\limits^\infty_0\frac{x cos \frac{1}{2} \pi x}{e^{2 \pi \sqrt{x}}-1} \ dx &=& \ds\frac{13-4 \pi}{8 \pi^2}, \\ \ds\int\limits^\infty_0 \frac{x \cos 2 \pi x}{e^{2 \pi \sqrt{x}}-1} \ dx &=& \frac{1}{64}\ds \left(\frac{1}{2} - \frac{3}{\pi} + \frac{5}{\pi^2}\right), \\ \ds\int\limits^\infty_0 \frac{x^2 \cos 2 \pi x}{e^{2 \pi \sqrt{x}}-1} \ dx &=& \frac{1}{256} \ds \left(1-\frac{5}{\pi}+\frac{5}{\pi^2}\right), \end{array}\right\} \end{eqnarray}
and so on.