Integrals of Rational Functions of Jacobi Elliptic Functions

Note that this post will be continuously updated as results are required for evaluating other integrals.

We use results from the relationships among squares of Jacobi elliptic functions.

Also, we use the numbering scheme of Handbook of Elliptic Integrals for Engineers and Scientists by Byrd and Friedman.

We begin with the definition of the incomplete elliptic integral of the third kind
\begin{equation}
\Pi(\phi,\alpha^{2},k) = \int \frac{du}{1 \,-\, \alpha^{2}\mathrm{sn}^{2}(u,k)}
\end{equation}

and
\begin{equation}
\int du = u
\end{equation}
which we note here as we drop the differential and dependent variables in the work below.

BF 337.01

We begin with
\begin{align}
\int \frac{\alpha^{2}\mathrm{sn}^{2}}{1 \,-\, \alpha^{2}\mathrm{sn}^{2}} &= \int \frac{1 \,-\, 1 + \alpha^{2}\mathrm{sn}^{2}}{1 \,-\, \alpha^{2}\mathrm{sn}^{2}} \\
&= \int \frac{1}{1 \,-\, \alpha^{2}\mathrm{sn}^{2}} \,-\, \int \frac{1 \,-\, \alpha^{2}\mathrm{sn}^{2}}{1 \,-\, \alpha^{2}\mathrm{sn}^{2}} = \Pi \,-\, u
\end{align}
Rearranging yields
\begin{equation}
\int \frac{\mathrm{sn}^{2}}{1 \,-\, \alpha^{2}\mathrm{sn}^{2}} = \frac{\Pi \,-\, u}{\alpha^{2}}
\end{equation}

BF 338.01

\begin{align}
\int \frac{\mathrm{cn}^{2}}{1 \,-\, \alpha^{2}\mathrm{sn}^{2}} &= \int \frac{1}{1 \,-\, \alpha^{2}\mathrm{sn}^{2}} \,-\, \int \frac{\mathrm{sn}^{2}}{1 \,-\, \alpha^{2}\mathrm{sn}^{2}} \\
&= \Pi \,-\, \frac{\Pi \,-\, u}{\alpha^{2}} = \frac{1}{\alpha^{2}}\left[(\alpha^{2} \,-\, 1)\Pi + u \right]
\end{align}

BF 339.01

\begin{align}
\int \frac{\mathrm{dn}^{2}}{1 \,-\, \alpha^{2}\mathrm{sn}^{2}} &= \int \frac{1 \,-\, k^{2}\mathrm{sn}^{2}}{1 \,-\, \alpha^{2}\mathrm{sn}^{2}} \\
&= \Pi \,-\, \frac{k^{2}}{\alpha^{2}}(\Pi \,-\, u) = \frac{1}{\alpha^{2}}\left[(\alpha^{2} \,-\, k^{2})\Pi + k^{2}u \right]
\end{align}

Integrals of Jacobi Elliptic Functions Raised to the Power 4

We use combinations of the following to evaluate integrals in this post:

All results can be found in section 310 – 321 of Handbook of Elliptic Integrals for Engineers and Scientists by Byrd and Friedman. We use their equation numbering scheme.

BF 310.04

\begin{align}
\int \mathrm{sn}^{4} &= \int \mathrm{sn}^{2} (1 \,-\, \mathrm{cn}^{2}) = \int \mathrm{sn}^{2} \,-\, \int \mathrm{sn}^{2} \mathrm{cn}^{2} \\
&= \frac{1}{3k^{4}}\left[(k^{2} + 2)u \,-\, 2(k^{2} + 1)E + k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} \right]
\end{align}
We used GR 5.134.1 and BF 361.01.

BF 312.04

\begin{align}
\int \mathrm{cn}^{4} &= \int \mathrm{cn}^{2} (1 \,-\, \mathrm{sn}^{2}) = \int \mathrm{cn}^{2} \,-\, \int \mathrm{sn}^{2} \mathrm{cn}^{2} \\
&= \frac{1}{3k^{4}}[{k^{\prime}}^{2}(2 \,-\, 3k^{2})u + 2(2k^{2} \,-\, 1)E + k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} ]
\end{align}
We used GR 5.134.2 and BF 361.01.

BF 314.04

\begin{align}
\int \mathrm{dn}^{4} &= \int \mathrm{dn}^{2} (1 \,-\, k^{2}\mathrm{sn}^{2}) = \int \mathrm{dn}^{2} \,-\, k^{2}\int \mathrm{sn}^{2} \mathrm{dn}^{2} \\
&= \frac{1}{3}[2(1 + {k^{\prime}}^{2})E \,-\, {k^{\prime}}^{2}u + k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} ]
\end{align}
We used GR 5.134.3 and BF 361.02.

Continue reading “Integrals of Jacobi Elliptic Functions Raised to the Power 4”

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}

Continue reading “Integrals of Gradshteyn and Ryzhik: 2.484 – Combinations of hyperbolic functions, exponentials, and powers”

Integrals of Pairs of Products of Squares of Jacobi Elliptic Functions

We use combinations of the following to evaluate integrals in this post:

The order of presentation here is done to avoid look ahead bias, thus avoiding circularity when referencing previous results.

All results can be found in section 361 of Handbook of Elliptic Integrals for Engineers and Scientists by Byrd and Friedman. We use their equation numbering scheme.

Note that this post will be continuously updated as results are required for evaluating other integrals.

BF 361.02

We begin with
\begin{equation}
\frac{\partial}{\partial u}\mathrm{cn} = -\,\mathrm{sn} \, \mathrm{dn}
\end{equation}
Squaring both sides, integrating by parts, and then making substitutions for \(\mathrm{cn}^{2}\), we have
\begin{align}
\int \mathrm{sn}^{2} \mathrm{dn}^{2} &= \int \left(\frac{\partial}{\partial u}\mathrm{cn}\right) \left(\frac{\partial}{\partial u}\mathrm{cn}\right) \\
&= -\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} + \int \mathrm{cn}^{2} \mathrm{dn}^{2} – k^{2}\int \mathrm{cn}^{2} \mathrm{sn}^{2} \\
&= -\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} + \int (1 – \,\mathrm{sn}^{2})\,\mathrm{dn}^{2} – \int (\mathrm{dn}^{2} – {k^{\prime}}^{2} ) \mathrm{sn}^{2} \\
&= \frac{1}{3 k^{2}}\left[{k^{\prime}}^{2}u + (2k^{2} – 1)E \,-\, k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} \right]
\end{align}

BF 361.01

From BF 361.02 we have
\begin{equation}
\int \mathrm{sn}^{2} ({k^{\prime}}^{2} + k^{2}\mathrm{cn}^{2}) = \frac{1}{3 k^{2}}\left[{k^{\prime}}^{2}u + (2k^{2} – 1)E \,-\, k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} \right]
\end{equation}
Rearranging yields
\begin{equation}
\int \mathrm{sn}^{2} \mathrm{cn}^{2} = \frac{1}{3 k^{4}}\left[(2 -\,k^{2})E \,-\,2{k^{\prime}}^{2}u \,-\, k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} \right]
\end{equation}

BF 361.03

From BF 361.02 we have
\begin{equation}
\int (1 \,-\,\mathrm{cn}^{2}) \mathrm{dn}^{2} = \frac{1}{3 k^{2}}\left[{k^{\prime}}^{2}u + (2k^{2} – 1)E \,-\, k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} \right]
\end{equation}
Rearranging yields
\begin{equation}
\int \mathrm{dn}^{2} \mathrm{cn}^{2} = \frac{1}{3 k^{2}}\left[(1+k^{2})E \,-\,{k^{\prime}}^{2}u + k^{2}\,\mathrm{cn} \,\mathrm{dn}\,\mathrm{sn} \right]
\end{equation}

Continue reading “Integrals of Pairs of Products of Squares of Jacobi Elliptic Functions”

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.