# Integrals of Gradshteyn and Ryzhik: 2.484 – Combinations of hyperbolic functions, exponentials, and powers

## Preliminary Work

We use the following definition of the exponential integral function

\mathrm{Ei}(z) = \int_{-\infty}^{z} \frac{\mathrm{e}^{x}}{x}dx

We also have,

\mathrm{Ei}(kz) = \int_{-\infty}^{kz} \frac{\mathrm{e}^{w}}{w}dw

which can be obtained by letting $$w=kx$$.

Additionally, we will require the following, which can be obtained via integration by parts

\label{eq:180831-1}
\tag{1}
\int \frac{\mathrm{e}^{kx}}{x^{2}}dx = -\,\frac{\mathrm{e}^{kx}}{x} + k\int \frac{\mathrm{e}^{kx}}{x} = k\,\mathrm{Ei}(kx)\,-\,\frac{\mathrm{e}^{kx}}{x}

For numbers 1-4 below, $$a^{2} \ne b^{2}$$, while for 5-10, $$a = b$$.

## 2.484.1

\begin{align}
\int \frac{1}{x}\mathrm{e}^{ax}\sinh(bx)\,dx &= \frac{1}{2} \int \frac{1}{x}[\mathrm{e}^{(a+b)x} \,-\, \mathrm{e}^{(a-b)x}] \\
&= \frac{1}{2} \left(\mathrm{Ei}[(a+b)x] \,-\, \mathrm{Ei}[(a-b)x]\right)
\end{align}

## 2.484.2

\begin{align}
\int \frac{1}{x}\mathrm{e}^{ax}\cosh(bx)\,dx &= \frac{1}{2} \int \frac{1}{x}[\mathrm{e}^{(a+b)x} + \mathrm{e}^{(a-b)x}] \\
&= \frac{1}{2} \left(\mathrm{Ei}[(a+b)x] + \mathrm{Ei}[(a-b)x]\right)
\end{align}

## 2.484.3

\begin{align}
\int \frac{1}{x^{2}}\mathrm{e}^{ax}\sinh(bx)\,dx &= \frac{1}{2} \int \frac{1}{x^{2}}[\mathrm{e}^{(a+b)x} \,-\, \mathrm{e}^{(a-b)x}] \\
&= \frac{1}{2} (a+b)\mathrm{Ei}[(a+b)x] \,-\, \frac{1}{2x}\mathrm{e}^{(a+b)x} \,-\, \frac{1}{2} (a-b)\mathrm{Ei}[(a-b)x] + \frac{1}{2x}\mathrm{e}^{(a-b)x} \\
&= \frac{1}{2} \left[(a+b)\mathrm{Ei}[(a+b)x] \,-\, (a-b)\mathrm{Ei}[(a-b)x]\right] \,-\, \frac{\mathrm{e}^{ax}}{x}\sinh(bx)
\end{align}
We used equation \eqref{eq:180831-1}. Also, there is an error in GR, an extra 2 in the denominator of the last term.

## 2.484.4

\begin{align}
\int \frac{1}{x^{2}}\mathrm{e}^{ax}\cosh(bx)\,dx &= \frac{1}{2} \int \frac{1}{x^{2}}[\mathrm{e}^{(a+b)x} + \mathrm{e}^{(a-b)x}] \\
&= \frac{1}{2} (a+b)\mathrm{Ei}[(a+b)x] \,-\, \frac{1}{2x}\mathrm{e}^{(a+b)x} + \frac{1}{2} (a-b)\mathrm{Ei}[(a-b)x] \,-\, \frac{1}{2x}\mathrm{e}^{(a-b)x} \\
&= \frac{1}{2} \left[(a+b)\mathrm{Ei}[(a+b)x] + (a-b)\mathrm{Ei}[(a-b)x]\right] \,-\, \frac{\mathrm{e}^{ax}}{x}\cosh(bx)
\end{align}
We used equation \eqref{eq:180831-1}. Also, there is an error in GR, an extra 2 in the denominator of the last term.

## 2.484.5

\begin{align}
\int \frac{1}{x}\mathrm{e}^{ax}\sinh(ax)\,dx &= \frac{1}{2} \int \frac{1}{x}[\mathrm{e}^{2ax} \,-\, 1] \\
&= \frac{1}{2} \left[\mathrm{Ei}(2ax) \,-\, \ln(x)\right]
\end{align}

## 2.484.6

\begin{align}
\int \frac{1}{x}\mathrm{e}^{-ax}\sinh(ax)\,dx &= \frac{1}{2} \int \frac{1}{x}[1 \,-\, \mathrm{e}^{-2ax}] \\
&= \frac{1}{2} \left[\ln(x) \,-\, \mathrm{Ei}(-2ax)\right]
\end{align}

## 2.484.7

\begin{align}
\int \frac{1}{x}\mathrm{e}^{ax}\cosh(ax)\,dx &= \frac{1}{2} \int \frac{1}{x}[\mathrm{e}^{2ax} + 1] \\
&= \frac{1}{2} \left[\mathrm{Ei}(2ax) + \ln(x)\right]
\end{align}

## 2.484.8

Let $$b = a$$ in GR 2.484.3.
\begin{align}
\int \frac{1}{x^{2}}\mathrm{e}^{ax}\sinh(ax)\,dx &= \frac{1}{2} \int \frac{1}{x^{2}}[\mathrm{e}^{2ax} \,-\, 1] \\
&= \frac{1}{2} 2a\mathrm{Ei}(2ax) \,-\, \frac{\mathrm{e}^{2ax}}{2x} + \frac{1}{2x} \\
&= a\,\mathrm{Ei}(2ax) \,-\, \frac{\mathrm{e}^{2ax}}{2x} + \frac{1}{2x}
\end{align}

## 2.484.9

Using the work done for GR 2.484.3 and GR 2.484.8, we have
\begin{align}
\int \frac{1}{x^{2}}\mathrm{e}^{-ax}\sinh(ax)\,dx &= \frac{1}{2} \int \frac{1}{x^{2}}[1 \,-\, \mathrm{e}^{-2ax}] \\
&= \,-\, \frac{1}{2x} + \frac{\mathrm{e}^{-2ax}}{2x} + \frac{1}{2} 2a\mathrm{Ei}(-2ax) \\
&= \,-\, \frac{1}{2x} + \frac{\mathrm{e}^{-2ax}}{2x} + a\,\mathrm{Ei}(-2ax)
\end{align}

## 2.484.10

Let $$b = a$$ in GR 2.484.4.
\begin{align}
\int \frac{1}{x^{2}}\mathrm{e}^{ax}\cosh(ax)\,dx &= \frac{1}{2} \int \frac{1}{x^{2}}[\mathrm{e}^{2ax} + 1] \\
&= \frac{1}{2} 2a\mathrm{Ei}(2ax) \,-\, \frac{\mathrm{e}^{2ax}}{2x} \,-\, \frac{1}{2x} \\
&= a\,\mathrm{Ei}(2ax) \,-\, \frac{\mathrm{e}^{2ax}}{2x} \,-\, \frac{1}{2x}
\end{align}