Integrals of Gradshteyn and Ryzhik: 5.136 – 5.139 – Integrals of Combinations of Jacobi Elliptic Functions (Part 1)

5.136.1

\begin{equation}
\int \mathrm{sn}\,\mathrm{cn} = -\frac{1}{k^2}\mathrm{dn}
\end{equation}

5.136.2

\begin{equation}
\int \mathrm{sn}\,\mathrm{dn} = -\mathrm{cn}
\end{equation}

5.136.3

\begin{equation}
\int \mathrm{cn}\,\mathrm{dn} = \mathrm{sn}
\end{equation}
The three integrals above follow from the derivatives of the Jacobi elliptic functions.

5.137.1

\begin{equation}
\int \frac{\mathrm{sn}}{\mathrm{cn}^2} = \frac{1}{{k^{\prime}}^2} \frac{\mathrm{dn}}{\mathrm{cn}} = \frac{1}{{k^{\prime}}^2} \mathrm{dc}
\end{equation}

5.137.2

\begin{equation}
\int \frac{\mathrm{sn}}{\mathrm{dn}^2} = -\frac{1}{{k^{\prime}}^2} \frac{\mathrm{cn}}{\mathrm{dn}}
= -\frac{1}{{k^{\prime}}^2} \mathrm{cd}
\end{equation}

5.137.3

\begin{equation}
\int \frac{\mathrm{cn}}{\mathrm{sn}^2} = -\frac{\mathrm{dn}}{\mathrm{sn}} = -\mathrm{ds}
\end{equation}

5.137.4

\begin{equation}
\int \frac{\mathrm{cn}}{\mathrm{dn}^2} = \frac{\mathrm{sn}}{\mathrm{dn}} = \mathrm{sd}
\end{equation}

5.137.5

\begin{equation}
\int \frac{\mathrm{dn}}{\mathrm{sn}^2} = -\frac{\mathrm{cn}}{\mathrm{sn}} = -\mathrm{cs}
\end{equation}

5.137.6

\begin{equation}
\int \frac{\mathrm{dn}}{\mathrm{cn}^2} = \frac{\mathrm{sn}}{\mathrm{cn}} = \mathrm{sc}
\end{equation}
The six integrals above follow from the derivatives of the Jacobi elliptic functions.

5.138.1

\begin{align}
\int \frac{\mathrm{cn}}{\mathrm{sn}\,\mathrm{dn}} &= \int \frac{\mathrm{cn}}{\mathrm{sn}\,\mathrm{dn}} \frac{\mathrm{dn}}{\mathrm{dn}} = \int \frac{\mathrm{cn}}{\mathrm{dn}^2} \frac{1}{\mathrm{sd}} \\
&= \int \frac{1}{w} = \ln(\mathrm{sd}) = \ln\left(\frac{\mathrm{sn}}{\mathrm{dn}}\right)
\end{align}
We used the substitution \(w=\mathrm{sd}\).

5.138.2

\begin{align}
\int \frac{\mathrm{sn}}{\mathrm{cn}\,\mathrm{dn}} &= \int \frac{\mathrm{sn}}{\mathrm{cn}\,\mathrm{dn}} \frac{\mathrm{cn}}{\mathrm{cn}} = \int \frac{\mathrm{sn}}{\mathrm{cn}^2} \frac{1}{\mathrm{dc}} \\
&= \frac{1}{{k^{\prime}}^2} \int \frac{1}{w} = \frac{1}{{k^{\prime}}^2} \ln(\mathrm{dc}) = \frac{1}{{k^{\prime}}^2} \ln\left(\frac{\mathrm{dn}}{\mathrm{cn}}\right)
\end{align}
We used the substitution \(w=\mathrm{dc}\).

5.138.3

\begin{align}
\int \frac{\mathrm{dn}}{\mathrm{sn}\,\mathrm{cn}} &= \int \frac{\mathrm{dn}}{\mathrm{sn}\,\mathrm{cn}} \frac{\mathrm{cn}}{\mathrm{cn}} = \int \frac{\mathrm{dn}}{\mathrm{cn}^2} \frac{1}{\mathrm{sc}} \\
&= \int \frac{1}{w} = \ln(\mathrm{sc}) = \ln\left(\frac{\mathrm{sn}}{\mathrm{cn}}\right)
\end{align}
We used the substitution \(w=\mathrm{sc}\).

5.139.1

\begin{equation}
\int \frac{\mathrm{cn}\,\mathrm{dn}}{\mathrm{sn}} = \int \frac{1}{w} = \ln(\mathrm{sn})
\end{equation}

5.139.2

\begin{align}
\int \frac{\mathrm{sn}\,\mathrm{dn}}{\mathrm{cn}} &= \int \frac{\mathrm{sn}\,\mathrm{dn}}{\mathrm{cn}} \frac{\mathrm{cn}}{\mathrm{cn}} = \int \frac{\mathrm{sn}\,\mathrm{dn}}{\mathrm{cn}^2} \frac{1}{\mathrm{nc}} \\
&= \int \frac{1}{w} = \ln(\mathrm{nc}) = \ln\left(\frac{1}{\mathrm{cn}}\right)
\end{align}
We used the substitution \(w=\mathrm{nc}\).

5.139.3

\begin{equation}
\int \frac{\mathrm{cn}\,\mathrm{sn}}{\mathrm{dn}} = -\frac{1}{k^2} \int \frac{1}{w} = -\frac{1}{k^2} \ln(\mathrm{dn})
\end{equation}

Integrals of Gradshteyn and Ryzhik: 3.267

We use the following definitions of the gamma and beta functions:
\begin{equation}
\Gamma(z) = \int\limits_{0}^{\infty} \mathrm{e}^{-t} t^{z-1} dt, \,\, \mathfrak{R}(z) \gt 0
\end{equation}

\begin{equation}
\mathrm{B}(a,b) = \int\limits_{0}^{1} t^{a-1} (1-t)^{b-1} dt
= \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}, \,\, \mathfrak{R}(a) \gt 0 \,\, \mathrm{and} \,\, \mathfrak{R}(b) \gt 0
\end{equation}

3.267.1

\begin{align}
\int\limits_{0}^{1} x^{3n} (1-x^3)^{-1/3} dx
&= \frac{1}{3} \int\limits_{0}^{1} y^{n-2/3} (1-y)^{-1/3} dy
= \frac{1}{3} \mathrm{B}\left( \frac{2}{3},n+\frac{1}{3} \right) \\
&= \frac{\Gamma\left(\frac{2}{3}\right) \Gamma\left(n + \frac{1}{3}\right)}{3\Gamma(n+1)}
= \frac{2\pi}{3\sqrt{3}} \frac{\Gamma\left(n + \frac{1}{3}\right)}{\Gamma\left(\frac{1}{3}\right) \Gamma(n+1)}
\end{align}

3.267.2

\begin{align}
\int\limits_{0}^{1} x^{3n-1} (1-x^3)^{-1/3} dx
&= \frac{1}{3} \int\limits_{0}^{1} y^{n-1} (1-y)^{-1/3} dy
= \frac{1}{3} \mathrm{B}\left(n, \frac{2}{3} \right) \\
&= \frac{\Gamma(n)\Gamma(2/3)}{3\Gamma\left(n + \frac{2}{3}\right)}
= \frac{(n-1)! \Gamma(2/3)}{3\Gamma\left(n + \frac{2}{3}\right)}
\end{align}

3.267.3

\begin{align}
\int\limits_{0}^{1} x^{3n-2} (1-x^3)^{-1/3} dx
&= \frac{1}{3} \int\limits_{0}^{1} y^{n-4/3} (1-y)^{-1/3} dy
= \frac{1}{3} \mathrm{B}\left( \frac{2}{3}, n \, – \frac{1}{3} \right) \\
&= \frac{\Gamma\left(n – \frac{1}{3}\right)\Gamma(2/3)}{3\Gamma\left(n + \frac{1}{3}\right)}
\end{align}

we used the substitution \(y=x^3\) for all three integrals.

Integrals of Gradshteyn and Ryzhik: 3.322

We define the error function as:
\begin{equation}
\mathrm{erf}(z) = \frac{2}{\sqrt{\pi}} \int\limits_{0}^{z} \mathrm{e}^{-x^2} dx
\end{equation}

3.322.1

\begin{align}
\int\limits_{u}^{\infty} \mathrm{e}^{-2ax-x^2} dx
&= \mathrm{e}^{a^2} \int\limits_{u}^{\infty} \mathrm{e}^{-(x+a)^2} dx
= \mathrm{e}^{a^2} \int\limits_{u+a}^{\infty} \mathrm{e}^{-y^2} dy \\
&= \frac{\sqrt{\pi}}{2} \mathrm{e}^{a^2} \mathrm{erf}(y) \Bigg|_{u+a}^{\infty} \\
&= \frac{\sqrt{\pi}}{2} \mathrm{e}^{a^2} \left[1-\,\mathrm{erf}(u+a)\right],\,\, u \gt 0
\end{align}
We completed the square in the argument of the exponential function and then used
the substitution \(y=x+a\).

3.322.2

\begin{equation}
\int\limits_{0}^{\infty} \mathrm{e}^{-2ax-x^2} dx =
\frac{\sqrt{\pi}}{2} \mathrm{e}^{a^2} \left[1-\,\mathrm{erf}(a)\right]
\end{equation}
follows from 3.322.1.

3.322.3

\begin{equation}
\mathrm{PV} \int\limits_{0}^{\infty} \mathrm{e}^{\pm i\lambda x^2} dx
= \frac{1}{\sqrt{\lambda}} \int\limits_{0}^{\infty} \mathrm{e}^{\pm i y^2} dy
= \frac{1}{2}\sqrt{\frac{\pi}{\lambda}} \mathrm{e}^{\pm i \pi /4}
\end{equation}
we let \(y^2=\lambda x^2\) and this problem was solved here.

Integrals of Gradshteyn and Ryzhik: 3.321

We define the error function as:
\begin{equation}
\mathrm{erf}(z) = \frac{2}{\sqrt{\pi}} \int\limits_{0}^{z} \mathrm{e}^{-x^2} dx
\end{equation}

3.321.2

\begin{equation}
\int\limits_{0}^{u} \mathrm{e}^{-q^{2}x^{2}} dx = \frac{1}{q} \int\limits_{0}^{qu} \mathrm{e}^{-z^{2}} dz
= \frac{\sqrt{\pi}}{2q} \mathrm{erf}(z) \Big|_{0}^{qu}
= \frac{\sqrt{\pi}}{2q} \mathrm{erf}(qu),\,\,q \gt 0
\end{equation}
using the substitution \(z=qx\)

3.321.3

\begin{equation}
\int\limits_{0}^{\infty} \mathrm{e}^{-q^{2}x^{2}} dx = \frac{\sqrt{\pi}}{2q},\,\,q \gt 0
\end{equation}
follows from 3.321.2

3.321.4

\begin{equation}
I = \int\limits_{0}^{u} x\mathrm{e}^{-q^{2}x^{2}} dx
= \frac{\sqrt{\pi}}{2q} \,x\, \mathrm{erf}(qx) \,- \frac{\sqrt{\pi}}{2q} \int \mathrm{erf}(qx) dx
\end{equation}
using integration by parts.
\begin{align}
\int \mathrm{erf}(qx) dx &= \frac{1}{q} \int \mathrm{erf}(y) dy \\
&= \frac{1}{q} \left[y\,\mathrm{erf}(y) – \frac{2}{\sqrt{\pi}} \int y\,\mathrm{e}^{-y^2} dy \right] \\
&= \frac{1}{q} y\,\mathrm{erf}(y) + \frac{1}{q\sqrt{\pi}} \mathrm{e}^{-y^2} \\
&= x\,\mathrm{erf}(qx) + \frac{1}{q\sqrt{\pi}} \mathrm{e}^{-q^{2}x^{2}}
\end{align}
using integration by parts and the substitution \(y=qx\)
\begin{equation}
I = \frac{\sqrt{\pi}}{2q} x \,\mathrm{erf}(qx) \,- \frac{\sqrt{\pi}}{2q} x \,\mathrm{erf}(qx)
\,- \frac{1}{2q^2} \mathrm{e}^{-q^{2}x^{2}} \Big|_{0}^{u}
= \frac{1}{2q^2} \left[1 – \mathrm{e}^{-q^{2}u^{2}} \right]
\end{equation}

3.321.5

\begin{align}
I &= \int\limits_{0}^{u} x^{2} \mathrm{e}^{-q^{2}x^{2}} dx
= \frac{-1}{2q^2} x \mathrm{e}^{-q^{2}x^{2}} + \frac{1}{2q^2} \int\mathrm{e}^{-q^{2}x^{2}} dx \\
&= \frac{1}{2q^3} \left[-qx \mathrm{e}^{-q^{2}x^{2}} + \frac{\sqrt{\pi}}{2} \mathrm{erf}(qx) \right]\Big|_{0}^{u} \\
&= \frac{1}{2q^3} \left[\frac{\sqrt{\pi}}{2} \mathrm{erf}(qu) \,-\, qu \mathrm{e}^{-q^{2}x^{2}} \right]
\end{align}
We used integration by parts, 3.321.4, and 3.321.2.

3.321.6

\begin{align}
I &= \int\limits_{0}^{u} x^{3} \mathrm{e}^{-q^{2}x^{2}} dx
= \frac{\sqrt{\pi}}{4q^3}x\,\mathrm{erf}(qx) \,- \frac{x^2}{2q^2}\mathrm{e}^{-q^{2}x^{2}}
– \frac{\sqrt{\pi}}{4q^3} \int \mathrm{erf}(qx) dx + \frac{1}{2q^2} \int x\,\mathrm{e}^{-q^{2}x^{2}} dx \\
&= \frac{\sqrt{\pi}}{4q^3}\,x\,\mathrm{erf}(qx) \,- \frac{x^2}{2q^2}\mathrm{e}^{-q^{2}x^{2}}
– \frac{\sqrt{\pi}}{4q^3}\,x\,\mathrm{erf}(qx) \,- \frac{1}{4q^4}\mathrm{e}^{-q^{2}x^{2}}
– \frac{1}{4q^4}\mathrm{e}^{-q^{2}x^{2}} \\
&= \frac{1}{2q^4} (-q^{2}x^{2} – 1)\mathrm{e}^{-q^{2}x^{2}}\Big|_{0}^{u} \\
&= \frac{1}{2q^4} \left[1-(1+q^{2}u^{2}) \mathrm{e}^{-q^{2}u^{2}} \right]
\end{align}
using integration by parts, 3.321.5, the work done in the solution of 3.321.4 and the result 3.321.4.

3.321.7

\begin{align}
I &= \int\limits_{0}^{u} x^{4} \mathrm{e}^{-q^{2}x^{2}} dx
= -\frac{1}{2q^4}x\mathrm{e}^{-q^{2}x^{2}} – \frac{1}{2q^4}q^{2}x^{3}\mathrm{e}^{-q^{2}x^{2}}
+ \frac{1}{2q^4}\int \mathrm{e}^{-q^{2}x^{2}} dx + \frac{q^{2}}{2q^4}\int x^{2}\mathrm{e}^{-q^{2}x^{2}} dx \\
&= \frac{1}{2q^5}\left[-qx\,\mathrm{e}^{-q^{2}x^{2}} – q^{3}x^{3}\,\mathrm{e}^{-q^{2}x^{2}}
+ \frac{\sqrt{\pi}}{2} \mathrm{erf}(qx) + \frac{\sqrt{\pi}}{4} \mathrm{erf}(qx)
-\frac{1}{2}qx\,\mathrm{e}^{-q^{2}x^{2}} \right]\Big|_{0}^{u} \\
&= \frac{1}{2q^5} \left[\frac{3\sqrt{\pi}}{4}\mathrm{erf}(qx)
– qx\,\mathrm{e}^{-q^{2}x^{2}}\left(\frac{3}{2} + q^{2}x^{2}\right) \right]\Big|_{0}^{u} \\
&= \frac{1}{2q^5} \left[\frac{3\sqrt{\pi}}{4}\mathrm{erf}(qu)
– qu\,\mathrm{e}^{-q^{2}u^{2}}\left(\frac{3}{2} + q^{2}u^{2}\right) \right]
\end{align}
using integration by parts, 3.321.6, 3.321.2, and 3.321.5.

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}

Evaluate the Integral \(\int_{0}^{\infty} \mathrm{e}^{-ax^2}\sin(wx) dx\)

This problem was posed at Mathematics Stack Exchange, here is my solution.

\begin{align}
I &= \int\limits_{0}^{\infty} \mathrm{e}^{-ax^2 + iwx} dx \\
\tag{a}
&= \mathrm{e}^{-\frac{w^2}{4a}} \int\limits_{0}^{\infty} \mathrm{e}^{-a(x-\frac{iw}{2a})^2} dx \\
&= \mathrm{e}^{-\frac{w^2}{4a}} \int \mathrm{e}^{-az^2} dz \\
&= \mathrm{e}^{-\frac{w^2}{4a}} \frac{1}{2}\sqrt{\frac{\pi}{a}} \mathrm{erf}(z\sqrt{a}) \\
&= \mathrm{e}^{-\frac{w^2}{4a}} \frac{1}{2}\sqrt{\frac{\pi}{a}}
\mathrm{erf}\left(x\sqrt{a} – \frac{iw}{2\sqrt{a}} \right) \Big|_{0}^{\infty} \\
\tag{b}
&= \mathrm{e}^{-\frac{w^2}{4a}} \frac{1}{2}\sqrt{\frac{\pi}{a}} \left[ 1 + i\,\mathrm{erfi}\left(\frac{w}{2\sqrt{a}} \right) \right]
\end{align}

a. Complete the square.

b. \(\lim_{x \to \infty} \mathrm{erf}(x + ic) = 1\) for finite \(c, \, c \in \mathbb{C}\) and \(\mathrm{erf}(-iz) = -i\mathrm{erfi}(z)\)

\begin{equation}
\int\limits_{0}^{\infty} \mathrm{e}^{-ax^2}\sin(wx) dx = \mathfrak{I}(I)
= \mathrm{e}^{-\frac{w^2}{4a}} \frac{1}{2}\sqrt{\frac{\pi}{a}} \mathrm{erfi}\left(\frac{w}{2\sqrt{a}} \right)
= \frac{1}{\sqrt{a}} \mathrm{F}\left(\frac{w}{2\sqrt{a}} \right)
\end{equation}

Where
\begin{equation}
\mathrm{F}(x) = \frac{\sqrt{\pi}}{2} \mathrm{e}^{-x} \mathrm{erfi}(x)
= \mathrm{e}^{-x^2} \int\limits_{0}^{x} \mathrm{e}^{z^2} dz
\end{equation}
is Dawson’s integral.

Evaluate the Integral \(\int_{0}^{1} \frac{\sqrt{1-x^{4}}}{1+x^{2}} dx\)

This problem was posed at Mathematics Stack Exchange, here is my solution.

Let \(z=x^{2}\)
\begin{align}
\int\limits_{0}^{1} \frac{\sqrt{1-x^{4}}}{1+x^{2}} dx &=
\int\limits_{0}^{1} (1-x^{2})^{1/2}(1+x^{2})^{-1/2} dx \\
&= \frac{1}{2} \int\limits_{0}^{1} (1-z)^{1/2} (1+z)^{-1/2} z^{-1/2} dz \\
&= \frac{\Gamma(1/2) \Gamma(3/2)}{2 \Gamma(2)} {}_{2}\mathrm{F}_{1}(1/2,1/2;2;-1) \\
&\approx 0.7119586
\end{align}

We used Gauss’s hypergeometric function.

Evaluate the Integral \(\int_{1}^{a^{2}} \frac{\ln x}{\sqrt{x}(x+a)} dx\)

This problem was posed at Mathematics Stack Exchange, here is my solution.

Let \(t=\sqrt{x}\)
\begin{equation}
I = \int\limits_{1}^{a^{2}} \frac{\ln x}{\sqrt{x}(x+a)} dx
= 4 \int\limits_{1}^{a} \frac{\ln t}{t^{2}+a} dt
\end{equation}

Integrating by parts, we have
\begin{align}
I_{1} &= \int\limits_{1}^{a} \frac{\ln t}{t^{2}+a} dt \\
&= \frac{\ln t}{\sqrt{a}} \tan^{-1}\left( \frac{t}{\sqrt{a}} \right) \Big|_{1}^{a}
\, – \frac{1}{\sqrt{a}} \int\limits_{1}^{a} \frac{1}{t} \tan^{-1}\left( \frac{t}{\sqrt{a}} \right) dt \\
&= \frac{\ln a}{\sqrt{a}} \tan^{-1}(\sqrt{a})
\, – \frac{1}{\sqrt{a}} \int\limits_{1}^{a} \frac{1}{t} \tan^{-1}\left( \frac{t}{\sqrt{a}} \right) dt
\end{align}

Let \(y=t/ \sqrt{a}\)
\begin{align}
I_{2} &= \int\limits_{1}^{a} \frac{1}{t} \tan^{-1}\left( \frac{t}{\sqrt{a}} \right) dt
= \int\limits_{1/\sqrt{a}}^{\sqrt{a}} \frac{1}{y} \tan^{-1}(y) dy \\
\tag{a}
&= \frac{i}{2} \left[ \int\limits_{1/\sqrt{a}}^{\sqrt{a}} \frac{\ln (1-iy)}{y} dy
\, – \int\limits_{1/\sqrt{a}}^{\sqrt{a}} \frac{\ln (1+iy)}{y} dy \right] \\
\tag{b}
&= \frac{i}{2} \left[ \mathrm{Li}_{2}(-iy) – \mathrm{Li}_{2}(iy) \right] \Big|_{1/\sqrt{a}}^{\sqrt{a}} \\
&= \frac{i}{2} \left( \left[ \mathrm{Li}_{2}(-i\sqrt{a}) + \mathrm{Li}_{2}\left(\frac{i}{\sqrt{a}}\right) \right]
– \left[ \mathrm{Li}_{2}(i\sqrt{a}) + \mathrm{Li}_{2}\left(\frac{-i}{\sqrt{a}}\right) \right] \right) \\
\tag{c}
&= \frac{i}{2} \left( \left[ -\frac{\pi ^{2}}{6} – \frac{1}{2} \ln ^{2}(i\sqrt{a}) \right]
– \left[ -\frac{\pi ^{2}}{6} – \frac{1}{2} \ln ^{2}(-i\sqrt{a}) \right] \right) \\
\tag{d}
&= \frac{\pi}{4} \ln a
\end{align}
a. \(\tan^{-1}(y) = \frac{i}{2} [\ln (1-iy) – \ln (1+iy)]\)

b. Dilogarithm function
\begin{equation}
\mathrm{Li}_{2}(z) = -\int_{0}^{z} \frac{\ln (1-x)}{x} dx
\end{equation}

c. Use the identity
\begin{equation}
\mathrm{Li}_{2}(z) + \mathrm{Li}_{2}(1/z) = -\frac{\pi ^{2}}{6} – \frac{1}{2} \ln ^{2}(-z)
\end{equation}

d. \(\ln (\pm iz) = \ln z \pm i\pi /2\)

Now we have
\begin{equation}
I = 4I_{1} = \frac{4}{\sqrt{a}} (\ln a) \tan^{-1}(\sqrt{a}) \, – \frac{\pi}{\sqrt{a}} \ln a
\end{equation}

Evaluate the Integral \(\int_{x_{0}}^{1} (1-x)^{a} x^{-a} dx\)

This problem was posed at Mathematics Stack Exchange, here is my solution.

\begin{align}
\int\limits_{x_{0}}^{1} (1-x)^{a} x^{-a} dx
&= \int\limits_{0}^{1} (1-x)^{a} x^{-a} dx – \int\limits_{0}^{x_{0}} (1-x)^{a} x^{-a} dx \\
&= \mathrm{B}(1-a,1+a) – \mathrm{B}_{x_{0}}(1-a,1+a) \\
&= \Gamma(1-a)\Gamma(1+a) – \frac{x_{0}^{1-a}}{1-a} \, {}_{2}\mathrm{F}_{1}(1-a,-1;2-a;x_{0})
\end{align}

We have used the incomplete beta function and Gauss’s hypergeometric function.

Evaluate the Integral \(\int \frac{x}{\mathrm{e}^{x}-1} dx\)

This question was posed at Mathematics Stack Exchange, here is my solution.

\begin{equation}
I = \int \frac{x}{\mathrm{e}^{x}-1} dx = -I_{1} = -\int \frac{x}{1-\mathrm{e}^{x}} dx
\end{equation}

Integrate by parts
\begin{align}
I_{1} &= \int \frac{x}{1-\mathrm{e}^{x}} dx \\
\tag{1}
&= x^{2} – x \ln(1-\mathrm{e}^{x}) – \int x dx + \int \ln(1-\mathrm{e}^{x}) dx \\
\tag{2}
&= x^{2} – x \ln(1-\mathrm{e}^{x}) – \frac{x^{2}}{2} – \mathrm{Li}_{2}(\mathrm{e}^{x})
\end{align}
and thus
\begin{equation}
\int \frac{x}{\mathrm{e}^{x}-1} dx = x \ln(1-\mathrm{e}^{x}) – \frac{x^{2}}{2} + \mathrm{Li}_{2}(\mathrm{e}^{x})
\end{equation}

1. Let \(u=\mathrm{e}^{x}\)
\begin{align}
\int \frac{1}{1-\mathrm{e}^{x}} dx &= \int \frac{1}{u(1-u)} du \\
&= \int \left( \frac{1}{u} + \frac{1}{1-u} \right) du \\
&= \ln \frac{u}{1-u} \\
&= x – \ln(1-\mathrm{e}^{x})
\end{align}

2. Let \(u=\mathrm{e}^{x}\)
\begin{align}
\int \ln(1-\mathrm{e}^{x}) dx &= \int \frac{\ln(1-u)}{u} du \\
&= -\mathrm{Li}_{2}(u) \\
&= -\mathrm{Li}_{2}(\mathrm{e}^{x})
\end{align}
where
\begin{equation}
\mathrm{Li}_{2}(z) = -\int\limits_{0}^{z} \frac{\ln(1-x)}{x} dx
\end{equation}
is the dilogarithm function.