Some definite integrals
Journal of the Indian Mathematical Society, XI, 1919, 81 – 87

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

$$\phi_w (t) = \frac{1}{\sqrt{w}} e^{-\frac{1}{4}\pi t^2 w'} \phi_{w'} (i t w').$$
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
$$e^{\frac{1}{4} \pi t^2/w}\{\half + \psi_w (t)\} = e^{\frac{1}{4} \pi(t+w)^{2/w}} \phi_w (t+w).$$
It is obvious that
$$\left. \begin{split} & \phi_w (t) = \phi_w (-t)\\ & \psi_w(t) = - \psi_w (-t) \end{split}\right\}.$$
From (1), (2) and (3) we easily find that
\begin{eqnarray} \half + \psi_w (t+i) = \frac{i}{\sqrt{w}} e^{-\frac{1}{4} \pi t^2/w} \left\{\half - \psi_{w'} \left(\frac{it}{w} + i\right)\right\}. \end{eqnarray}
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
$$\phi_w (t+i) + \phi_w(t-i) = \frac{1}{\sqrt{w}} e^{-\frac{1}{4}\pi t^2/ w};$$
and
\begin{eqnarray} \psi_w (t+i) - \psi_w(t-i) = \frac{i}{\sqrt{w}} e^{-\frac{1}{4}\pi t^2/ w}. \end{eqnarray}
From (1) and (5) we deduce that
$$e^{\frac{1}{4}\pi(t+w)^2/ w} \phi_w (t+w) + e^{\frac{1}{4}\pi(t- w)^2/w} \phi_w(t-w) = e^{\frac{1}{4}\pi t^2/w}.$$
Similarly from (4) and (6) we obtain
$$e^{\frac{1}{4}\pi(t+w)^2/ w} \left\{\half-\psi_w (t+w)\right\} = e^{\frac{1}{4}\pi(t- w)^2/w} \{\half + \psi_w (t-w)\}.$$
It is easy to deduce from (5) that if $n$ is a positive integer, then
\begin{align} & \phi_w(t)+(-1)^{n+1} \phi_w (t \pm 2ni)\notag\\ &\qquad = \frac{1}{\sqrt{w}} \left\{e^{-\frac{1}{4}\pi (t\pm i)^2/w} - e^{-\frac{1}{4}\pi (t\pm 3i)^2/w} + e^{-\frac{1}{4}\pi (t\pm 5i)^2/w} - \cdots \:{\rm to}\: n \:{\rm terms}\:\right\}. \end{align}
Similarly from (6) we have
\begin{align} & \psi_w(t) - \psi_w (t \pm 2ni)\notag\\ &\qquad = \mp \frac{i}{\sqrt{w}} \left\{e^{-\frac{1}{4}\pi (t + i)^2/w} + e^{-\frac{1}{4}\pi (t + 3i)^2/w} + e^{-\frac{1}{4}\pi (t + 5i)^2/w} + \cdots \:{\rm to}\: n \:{\rm terms}\:\right\}. \end{align}
Again from (7) we have
\begin{eqnarray} && e^{\frac{1}{4}\pi t^2/w} \phi_w (t) + (-1)^{n+1} e^{\frac{1}{4}\pi(t+2nw)^2/w \phi_w(t+2nw)}\notag\\ &&\qquad = e^{\frac{1}{4} \pi(t+w)^2/w} - e^{\frac{1}{4}\pi(t+3w)^2/w} + e^{\frac{1}{4}\pi(t+5w)^2/w} - \cdots \:\:{\rm to}\: n \:{\rm terms}; \end{eqnarray}
and from (8)
\begin{eqnarray} e^{\frac{1}{4} \pi t^2/w} \{\half + \psi_w (t) \} + (-1)^{n+1} e^{\frac{1}{4}\pi(t+2nw)^2/w} \{\half + \psi_w (t+2nw)\}\notag\\ = e^{\frac{1}{4} \pi(t+2w)^2/w} - e^{\frac{1}{4}\pi(t+4w)^2/w} + e^{\frac{1}{4}\pi(t+6w)^2/w} - \cdots \:\:{\rm to}\: n \:{\rm terms}. \end{eqnarray}

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

\begin{align} & \phi_w(s) + (-1)^{(m+1)(n+1)} e^{-\half \pi m (s+t)} \phi_w(t)\notag\\ &\quad = e^{-\frac{1}{4} \pi s^2/w} \left\{e^{\frac{1}{4} \pi (s-w)^2/w} - e^{\frac{1}{4} \pi (s-3w)^2/w} + e^{\frac{1}{4} \pi (s-5w)^2/w} - \cdots \:{\rm to}\: m \:{\rm terms} \:\right\}\notag\\ &\qquad + \frac{(-1)^{(m+1)(n+1)}}{\sqrt{w}} e^{-\half\pi m(s+t)} \notag\\ &\qquad \times \left\{e^{-\frac{1}{4}\pi(t\pm i)^2/w} - e^{-\frac{1}{4}\pi(t\pm 3i)^2/w} + e^{-\frac{1}{4}\pi(t\pm 5i)^2/w} - \cdots \:{\rm to}\: n \:{\rm terms}\:\right\}. \end{align}
Similarly, combining (10) and (12), we obtain
\begin{align} & \half-\psi_w(s) + (-1)^{mn+m+1} e^{-\half \pi m (s+t)}\{\half -\psi_w(t)\} \notag\\ &\quad = e^{-\frac{1}{4} \pi s^2/w} \left\{e^{\frac{1}{4} \pi (s-2w)^2/w} - e^{\frac{1}{4} \pi (s-4w)^2/w} + e^{\frac{1}{4} \pi (s-6w)^2/w} - \cdots \:{\rm to}\: m \:{\rm terms} \:\right\}\notag\\ &\qquad \pm (-1)^{mn+m+1} \frac{i}{\sqrt{w}} e^{-\half \pi m (s+t)}\notag\\ &\qquad \times \left\{e^{-\frac{1}{4}\pi(t\pm i)^2/w}+ e^{-\frac{1}{4}\pi(t\pm 3i)^2/w} + e^{-\frac{1}{4}\pi(t\pm 5i)^2/w} + \cdots \:{\rm to}\: n \:{\rm terms}\:\right\}, \end{align}
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

\begin{align} & \phi_w(t) \{1 + (-1)^{(m+1)(n+1)} e^{-\pi m t}\}\notag\\ &\quad = e^{-\frac{1}{4} \pi t^2/w} \left\{e^{\frac{1}{4} \pi (t-w)^2/w} - e^{\frac{1}{4} \pi (t-3w)^2/w} + e^{\frac{1}{4} \pi (t-5w)^2/w} - \cdots \:{\rm to}\:m \:{\rm terms} \:\right\}\notag\\ &\qquad + \frac{(-1)^{(m+1)(n+1)}}{\sqrt{w}} e^{-\pi mt} \left\{e^{-\frac{1}{4}\pi(t-i)^2/w} - e^{-\frac{1}{4}\pi(t-3i)^2/w} + \cdots \:{\rm to}\: n \:{\rm terms}\:\right\}; \end{align}
\begin{align} & \{\half-\psi_w(t)\} \{1 + (-1)^{mn+m+1} e^{-\pi m t}\}\notag\\ &\quad \quad = e^{-\frac{1}{4} \pi t^2/w} \left\{e^{\frac{1}{4} \pi (t-2w)^2/w} -e^{\frac{1}{4} \pi (t-4w)^2/w} + \cdots \: {\rm to}\: m \:{\rm terms}\right\}\notag\\ &\qquad + (-1)^{mn+m} \frac{i}{\sqrt{w}} e^{-\pi mt} \left\{e^{-\frac{1}{4}\pi(t-i)^2/w} - e^{-\frac{1}{4}\pi(t-3i)^2/w} + \cdots \:{\rm to}\: n \:{\rm terms}\:\right\}. \end{align}
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

\begin{align} & 2 \cosh nt \int^\infty_0 \frac{\cos 2tx}{\cosh \pi x} \cos \left(\frac{\pi m x^2}{n}\right) dx \notag\\ &\quad = [\cosh \{(1-n) t\} \cos (\pi m/4n) - \cosh \{(3-n)t\} \cos(9 \pi m/4n) + \cdots \:{\rm to}\: n \:{\rm terms}]\notag\\ &\qquad +\: \sqrt{\left(\frac{n}{m}\right)} \left[\cosh \left\{\left( 1-\frac{1}{m}\right)nt\right\} \cos \left(\frac{\pi}{4} - \frac{nt^2}{\pi m} + \frac{\pi n}{4m}\right) \right.\notag\\ &\qquad \left. - \cosh \left\{\left(1-\frac{3}{m}\right)nt\right\} \cos \left(\frac{\pi}{4} -\frac{nt^2}{\pi m} + \frac{9 \pi n}{4m}\right) + \cdots \:{\rm to}\: m \:{\rm terms}\right]; \end{align}
and
\begin{align} & 2 \cosh nt \int^\infty_0 \frac{\cos 2tx}{\cosh \pi x} \sin \left(\frac{\pi m x^2}{n}\right) dx \notag\\ &\quad = -[\cosh \{(1-n) t\} \sin (\pi m/4n) - \cosh \{(3-n)t\} \sin(9 \pi m/4n)\notag\\ &\qquad + \cosh \{(5-n)t\} \sin (25 \pi/4n) - \cdots \:{\rm to}\: n\:{\rm terms}]\notag\\ &\qquad + \sqrt{\left(\frac{n}{m}\right)} \left[\cosh \left\{\left( 1-\frac{1}{m} \right)nt\right\}\sin\left(\frac{\pi}{4} - \frac{nt^2}{\pi m}+\frac{\pi n}{4m}\right) \right.\notag\\ &\qquad \left. -\cosh\left\{\left(1-\frac{3}{m}\right)nt\right\} \sin\left(\frac{\pi}{4} -\frac{nt^2}{\pi m} + \frac{9 \pi n}{\pi m}\right) + \cdots \:{\rm to}\: n \:{\rm terms}\right]. \end{align}

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 formulĂ¦ 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

\begin{eqnarray} \phi_w(t) = \half e^{-\frac{1}{4} \pi t^2/w} \left\{e^{\frac{1}{4} \pi (t-w)^2/w} - e^{\frac{1}{4} \pi (t-3w)^2/w} + \cdots \:{\rm to}\: m \:{\rm terms} \right\} \notag\\ + \frac{1}{2 \sqrt{w}} \left\{e^{-\frac{1}{4}\pi (t\mp i)^2/w} - e^{-\frac{1}{4}\pi (t\mp 3i)^2/w}+\cdots\:{\rm to}\: n \:{\rm terms}\right\}. \end{eqnarray}

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

\begin{align} & \phi_w(t) + (-1)^{(1+\half m)(1+\half n)} e^{-\frac{1}{4} \pi mt} \phi_w (0) \notag\\ &\quad = e^{-\frac{1}{4}\pi t^2/w} \left\{e^{\frac{1}{4}\pi(t-w)^2/w} - e^{\frac{1}{4}\pi(t-3w)^2/w} + \cdots \:{\rm to}\:\: \half m \:{\rm terms}\right\} \notag\\ &\qquad + \frac{(-1)^{1+\half m)(1+\half n)}}{\sqrt{w}} e^{-\frac{1}{4} \pi mt} \left\{e^{\frac{1}{4}\pi/w} - e^{\frac{9}{4} \pi/w} + e^{\frac{25}{4} \pi/w} - \cdots {\rm to}\:\:\half n \:{\rm terms}\right\}. \end{align}
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
\begin{align} \psi_w(t) &= -\half e^{-\frac{1}{4}\pi t^2/w} \left\{e^{\frac{1}{4} \pi(t-2w)^2/w} - e^{\frac{1}{4}\pi(t-4w)^2/w} + \cdots \:{\rm to}\:m\:{\rm terms}\right\}\notag\\ &\quad \pm \frac{i}{2\sqrt{w}} \left\{e^{-\frac{1}{4}\pi(t\mp i)^2/w} + e^{-\frac{1}{4}\pi (t\mp3i)^2/w} + \cdots \:{\rm to} \:n \:{\rm terms}\right\}. \end{align}
If $t = mw \pm ni$, where $m$ is odd and $n$ is even, then
\begin{align} & \half - \psi_w(t) + \{(-1)^{1+\frac{1}{4}(m-1)(n+2)} e^{-\frac{1}{4}\pi\{(m-1)t+mw\}} \phi_w (0) \notag\\ &\quad = e^{-\frac{1}{4} \pi t^2/w} \left\{e^{\frac{1}{4}\pi (t-2w)^2/w} - e^{\frac{1}{4} \pi(t-4w)^2/w} \cdots + \:{\rm to}\:\half (m-1) \:{\rm terms}\right\}\notag\\ &\qquad \pm (-1)^{1+\frac{1}{4} (m-1)(n+2)} \frac{i}{\sqrt{w}} e^{-\frac{1}{4}\pi(m-1)(t+w)}\notag\\ &\qquad \times \left\{e^{-\frac{1}{4}\pi (w\pm i)^2/w} - e^{-\frac{1}{4}\pi(w \pm 3i)^2/w} + \cdots \:{\rm to} \:\:\half n \:{\rm terms}\right\}. \end{align}
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 formulĂ¦: \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.

Endnotes

1Messenger of Mathematics, Vol,44, 1915, pp. 75 – 85 [No.12 of this volume].