## 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}

## Evaluate the Integral $$\int_{0}^{1} \frac{1}{1-\log_{2}x} dx$$

How to evaluate

\int\limits_{0}^{1} \frac{1}{1-\log_{2}x} dx

was a question posed at Mathematics Stack Exchange. Here is my solution.

Make successive substitutions $$z=1-\frac{\ln x}{\ln2}, \,y=z\ln2$$

\begin{align}
\int\limits_{0}^{1} \frac{1}{1-\log_{2}x} dx &= 2\ln2 \int\limits_{1}^{\infty} \frac{1}{z} \mathrm{e}^{-z\ln2} dz \\
&= 2\ln2 \int\limits_{\ln2}^{\infty} \frac{\mathrm{e}^{-y}}{y} dy \\
&=-(2\ln2)\mathrm{Ei}(-\ln2) \approx 0.52495
\end{align}

\mathrm{Ei}(z) = -\int\limits_{-z}^{\infty} \frac{\mathrm{e}^{-t}}{t} dt

is the exponential integral function.