I have shewn elsewhere¹ that the definite integrals \begin{align*} \phi_w (t) & = \int\limits^\infty_0 \frac{\cos \pi t x}{\cosh \pi x} e^{-\pi w x^2} dx,\\ \psi_w (t) & = \int\limits^\infty_0 \frac{\sin \pi t x}{\sinh \pi x} e^{-\pi w x^2} dx \end{align*} can be evaluated in finite terms if $w$ is any rational multiple of $i$.

In this paper I shall shew, by a much simpler method, that these integrals can be evaluated not only for these values but also for many other values of $t$ and $w$.

Now we have \begin{eqnarray*} \phi_w(t) &=& 2 \int^\infty_0 \int^\infty_0 \frac{\cos 2 \pi x z}{\cosh \pi z} \cos \pi t x e^{-\pi w x^2} dx dz \\ &=& \frac{e^{-\frac{1}{4} \pi t^2 w'}}{\sqrt{w}} \int^\infty_0 \frac{\cosh \pi tx w'}{\cosh \pi x} e^{-\pi x^2 w'} dx \end{eqnarray*} where $w'$ stands for $1/w$.

It follows that

Again \begin{eqnarray*} \phi_w (t+w) &=& \frac{1}{\sqrt{w}} e^{-\frac{1}{4} \pi(t+w)^2 w'} \\ && \times \int^\infty_0 \frac{\cosh(\pi t x/w) \cosh \pi x + \sinh \pi t x/w \sinh \pi x}{\cosh \pi x} e^{-\pi x^2/w} dx \\ &=& \frac{1}{\sqrt{w}} e^{-\frac{1}{4} \pi(t+w)^2/w} \\ && \times \left\{\half \sqrt{w} e^{\frac{1}{4} \pi t^2/w} + 2 \int^\infty_0 \int^\infty_0 \frac{\sin 2 \pi x z}{\sinh \pi z} \sinh \frac{\pi t x}{w} e^{-\pi x^2/w} dx dz \right\} \end{eqnarray*} \begin{eqnarray*} &=& \frac{1}{\sqrt{w}} e^{-\frac{1}{4} \pi(t+w)^2/w} \\ && \times \left\{\half \sqrt{w} e^{\frac{1}{4} \pi t^2/w} +\sqrt{w} e^{\frac{1}{4} \pi t^2/w} \int^\infty_0 \frac{\sin \pi t x}{\sinh \pi x} e^{-\pi w x^2} dx \right\}. \end{eqnarray*} In other words It is obvious that From (1), (2) and (3) we easily find that It is easy to see that \begin{eqnarray*} \phi_w (i) = \frac{1}{2\sqrt{w}}; \;\;\; \psi_w(i) = \frac{i}{2\sqrt{w}}; \;\;\; \phi_w(w) = \half e^{-\frac{1}{4}\pi w}; \\ \half - \psi_w (w) =e^{-\frac{1}{4}\pi w} \phi_w (0); \;\;\; \phi_w (w \pm i) = \left(\frac{1}{2\sqrt{w}} + \frac{i}{2}\right) e^{-\frac{1}{4}\pi w}; \\ \psi_w (w\pm i) = \half \pm \frac{i}{2\sqrt{w}} e^{-\frac{1}{4}\pi w};\;\;\; \phi_w(\half w) + \psi_w(\half w) = \half. \end{eqnarray*} Again we see that and From (1) and (5) we deduce that Similarly from (4) and (6) we obtain It is easy to deduce from (5) that if $n$ is a positive integer, then Similarly from (6) we have Again from (7) we have and from (8)

Now, combining (9) and (11), we deduce that, if $m$ and $n$ are positive integers and $s=t+2mw \pm 2ni$, then

Similarly, combining (10) and (12), we obtain where $s$ and $t$ have the same relation as in (13).

Suppose now that $s=t$ in (13) and (14). Then we see that, if $w = in/m$, then

where $\sqrt{w}$ should be taken as $$e^{\frac{1}{4} \pi i} \sqrt{ \left(\frac{n}{m}\right)}.$$

In (15) and (16) there is no loss of generality in supposing that one of the two numbers $m$ and $n$ is odd.

Now equating the real and imaginary parts in (15), we deduce that, if $m$ and $n$ are positive integers of which one is odd, then

and

Equating the real and imaginary parts in (16), we can find similar expressions for the integrals $$ \int^\infty_0 \frac{\sin tx}{\sinh \pi x} \sin \left(\frac{\pi m x^2}{n}\right) dx, \int^\infty_0 \frac{\sin tx}{\sinh \pi x} \cos \left(\frac{\pi m x^2}{n}\right) dx. $$

From these formulæ we can evaluate a number of definite integrals, such as \begin{eqnarray*} \int^\infty_0 \frac{\cos 2 \pi t x}{\cosh \pi x} \cos \pi x^2 dx &=& \frac{1 +\sqrt{2} \sin \pi t^2}{2 \sqrt{2} \cosh \pi t}, \\ \int^\infty_0 \frac{\cos 2 \pi t x}{\cosh \pi x} \sin \pi x^2 dx &=& \frac{-1 + \sqrt{2} \cos \pi t^2}{2 \sqrt{2} \cosh \pi t}, \\ \int^\infty_0 \frac{\sin 2 \pi t x}{\sinh \pi x} \cos \pi x^2 dx &=& \frac{\cosh \pi t - \cos \pi t^2}{2 \sinh \pi t}, \\ \int^\infty_0 \frac{\sin 2 \pi t x}{\sinh \pi x} \sin \pi x^2 dx &=& \frac{\sin \pi t^2}{2 \sinh \pi t}, \end{eqnarray*} and so on.

Again supposing that $s=-t$ in (13), we deduce that if $t = mw \pm ni$, where $m$ and $n$ are positive integers of which one at least is odd, then

This formula is not true when both $m$ and $n$ are even.

If $t = mw \pm ni$, where $m$ and $n$ are both even, then

This is easily obtained by putting $t=0$ and then changing $s$ to $t$ in (13). Similarly from (14) we deduce that if $t = mw \pm ni$, where $m$ and $n$ are both even, or both odd, or $m$ is even and $n$ is odd, then If $t = mw \pm ni$, where $m$ is odd and $n$ is even, then This is obtained by putting $t=w$ in (14). A number of definite integrals such as the following can be evaluated with the help of the above formulæ: \begin{align*} \int^\infty_0 \frac{\cos \pi tx}{\cosh \pi x} e^{-\pi(t+i)x^2} dx &= \frac{1+i}{2\sqrt{2}} e^{-\frac{1}{4}\pi t} \left\{1-\frac{i}{\sqrt{(t+i)}} \right\},\\ \int^\infty_0 \frac{\sin \pi t x}{\sinh \pi x} e^{-\pi(t+i)x^2} dx &= \frac{1}{2} - \frac{1+i}{2\sqrt{2}} \cdot \frac{e^{-\frac{1}{4} \pi t} }{\sqrt{(t+i)}}, \end{align*} and so on.