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
\begin{equation}
\mathrm{Ei}(z) = \int_{-\infty}^{z} \frac{\mathrm{e}^{x}}{x}dx
\end{equation}
We also have,
\begin{equation}
\mathrm{Ei}(kz) = \int_{-\infty}^{kz} \frac{\mathrm{e}^{w}}{w}dw
\end{equation}
which can be obtained by letting \(w=kx\).

Additionally, we will require the following, which can be obtained via integration by parts
\begin{equation}
\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}
\end{equation}

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}

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

How to evaluate
\begin{equation}
\int\limits_{0}^{1} \frac{1}{1-\log_{2}x} dx
\end{equation}
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}

\begin{equation}
\mathrm{Ei}(z) = -\int\limits_{-z}^{\infty} \frac{\mathrm{e}^{-t}}{t} dt
\end{equation}
is the exponential integral function.