Evaluate the Integral \(\int_{0}^{1} \mathrm{li}(x) dx\)

\begin{equation}
\int\limits_{0}^{1} \mathrm{li}(x) dx = -\ln 2
\end{equation}
is entry 6.211 in Gradshteyn and Ryzhik’s Table of Integrals, Series, and Products.

\begin{equation}
\mathrm{li}(x) = \int\limits_{0}^{x} \frac{dt}{\ln t},\,\,|\mathrm{arg}x| \lt \pi ,\,\,|\mathrm{arg}(1-x)| \lt \pi
\end{equation}
is the logarithmic integral function.

We begin with the confluent hypergeometric function of the second kind (also known as
Tricomi’s confluent hypergeometric function)
\begin{equation}
\Psi(a,b;z) = \frac{1}{\Gamma(a)} \int\limits_{0}^{\infty} \mathrm{e}^{-zt} t^{a-1} (1+t)^{b-a-1} dt,
\,\, \mathfrak{R}(a) \gt 0,\,\, \mathfrak{R}(z) \gt 0
\end{equation}

We can express the logarithmic integral function in terms of the confluent hypergeometric function of the second kind
\begin{equation}
\mathrm{li}(x) = -x\Psi(1,1;-\ln x) = \int\limits_{0}^{\infty} \frac{x^t}{t+1} dt
\end{equation}

\begin{align}
\int\limits_{0}^{1} \mathrm{li}(x) dx &= -\int\limits_{0}^{1} \int\limits_{0}^{\infty} \frac{x^{t+1}}{t+1} dt dx \\
&= -\int\limits_{0}^{\infty} \frac{1}{t+1} \int\limits_{0}^{1} x^{t+1} dx dt \\
&= -\int\limits_{0}^{\infty} \frac{1}{(t+1)(t+2)} dt \\
&= [\ln(t+2)-\ln(t+1)]\Big|_{0}^{\infty} \\
&= -\ln 2
\end{align}