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

\int\limits_{0}^{1} \mathrm{li}(x) dx = -\ln 2

is entry 6.211 in Gradshteyn and Ryzhik’s Table of Integrals, Series, and Products.

\mathrm{li}(x) = \int\limits_{0}^{x} \frac{dt}{\ln t},\,\,|\mathrm{arg}x| \lt \pi ,\,\,|\mathrm{arg}(1-x)| \lt \pi

is the logarithmic integral function.

We begin with the confluent hypergeometric function of the second kind (also known as
Tricomi’s confluent hypergeometric function)

\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

We can express the logarithmic integral function in terms of the confluent hypergeometric function of the second kind

\mathrm{li}(x) = -x\Psi(1,1;-\ln x) = \int\limits_{0}^{\infty} \frac{x^t}{t+1} dt

\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 $$\mathrm{erf}(-iz) = -i\mathrm{erfi}(z)$$

\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)

Where

\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

is Dawson’s integral.

## Legendre-type relations for generalized complete elliptic integrals by Shingo Takeuchi

Legendre’s relation for the complete elliptic integrals of the first and second kinds is generalized. The proof depends on an application of the generalized trigonometric functions and is alternative to the proof for Elliott’s identity.

The entire paper is available here.

## 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 $$t=\sqrt{x}$$

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

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

\mathrm{Li}_{2}(z) = -\int_{0}^{z} \frac{\ln (1-x)}{x} dx

c. Use the identity

\mathrm{Li}_{2}(z) + \mathrm{Li}_{2}(1/z) = -\frac{\pi ^{2}}{6} – \frac{1}{2} \ln ^{2}(-z)

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

Now we have

I = 4I_{1} = \frac{4}{\sqrt{a}} (\ln a) \tan^{-1}(\sqrt{a}) \, – \frac{\pi}{\sqrt{a}} \ln a

## On The Extended Incomplete Pochhammer Symbols and Hypergeometric Functions by Rakesh Kumar Parmar, R.K. Raina

In this paper, we first introduce certain forms of extended incomplete Pochhammer symbols which are then used to define families of extended incomplete generalized hypergeometric functions. For these functions, we investigate various properties including the integral representations, derivative formula, certain generating function and fractional integrals (and derivatives) relationships. Some special cases of the main results are also deduced.

The entire paper is available here.

## A Collection of Free Math Books from Folkscanomy Mathematics: Books of a Mathematic Nature

See Folkscanomy Mathematics: Books of a Mathematic Nature. Here, you can find such books as The Theory of Functions by E.C. Titchmarsh and A Treatise on the Theory of Bessel Functions by G.N. Watson.

## Some formulae for products of Fubini polynomials with applications by Levent Kargın

In this paper we evaluate sums and integrals of products of Fubini polynomials and have new explicit formulas for Fubini polynomials and numbers. As a consequence of these results new explicit formulas for p-Bernoulli numbers and Apostol-Bernoulli functions are given. Besides, integrals of products of Apostol-Bernoulli functions are derived.

The entire paper is available here.

## Higher order generalized geometric polynomials by Levent Kargin, Bayram Çekim

According to generalized Mellin derivative (Kargin), we introduce a new family of polynomials called higher order generalized geometric polynomials. We obtain some properties of them.We discuss their connections to degenerate Bernoulli and Euler polynomials. Furthermore, we find new formulas for the Carlitz’s (Carlitz) and Howard’s (Howard2) finite sums. Finally, we evaluate several series in closed forms, one of which has the coefficients include values of the Riemann zeta function. Moreover, we calculate some integrals in terms of generalized geometric polynomials.

The entire paper is available here.

## Introduction

The Gudermannian function, named in honor of Christoph Gudermann, is a way to relate the trigonometric and hyperbolic functions without the use of complex variables. The basic definition is

\mathrm{gd} u = \int\limits_{0}^{u} \frac{1}{\cosh x} dx
\tag{1}
\label{eq:g1}

## Derivation of the Basic Definition and Basic Properties

To derive equation \eqref{eq:g1}, we begin with the hyperbola $$x^{2} – y^{2} = a^{2}$$. A point on the hyperbola has coordinates $$(x,y)$$ where

x = a \sec\theta = a \cosh u \quad \mathrm{and} \quad y = a \tan\theta = a \sinh u

Thus we have:

Using basic trigonometry we obtain

\cot\theta = \mathrm{csch} u \,\, \cos\theta = \mathrm{sech} u \,\, \sin\theta = \tanh u \,\,
\csc\theta = \coth u

We solve $$\tan\theta = \sinh u$$ for $$\theta$$ to obtain, $$\theta = \tan^{-1}(\sinh u)$$. Differentiating yields

\frac{d\theta}{du} = \frac{\cosh u}{1 + \sinh^{2} u} = \frac{1}{\cosh u}

Integrating yields

\theta = \int_{0}^{u} \frac{1}{\cosh t} dt

If we designate $$\theta$$ as the Gudermannian of $$u$$, $$\theta = \mathrm{gd} u$$, we obtain Equation \eqref{eq:g1}. Also, we can rewrite our relations between the trig and hyperbolic functions as
\begin{align}
\sec(\mathrm{gd} u) &= \cosh u, \quad \tan(\mathrm{gd} u) = \sinh u, \quad \cot(\mathrm{gd} u) = \mathrm{csch} u \\
\cos(\mathrm{gd} u) &= \mathrm{sech} u, \quad \sin(\mathrm{gd} u) = \tanh u, \quad \csc(\mathrm{gd} u) = \coth u
\end{align}

1. Let $$y = \tanh z/2$$
\begin{align}
\frac{1}{2} \mathrm{gd} u &= \frac{1}{2} \int\limits_{0}^{u} \frac{1}{\cosh z} dz
= \int\limits_{0}^{u} \frac{1}{\mathrm{e}^{z}+\mathrm{e}^{-z}} dz \\
&= \int\limits_{0}^{\tanh u/2} \frac{1}{y^{2}+1} dy \\
&= \tan^{-1}\left(\tanh\frac{u}{2} \right)
\end{align}
We can rewrite this as

\tan\left(\frac{1}{2} \mathrm{gd} u \right) = \tanh\frac{u}{2}

2. Let $$y= \mathrm{e}^{z}$$
\begin{align}
\mathrm{gd} u &= 2\int\limits_{0}^{u} \frac{1}{\mathrm{e}^{z}+\mathrm{e}^{-z}} dz
= 2\int\limits_{0}^{u} \frac{\mathrm{e}^{z}}{\mathrm{e}^{2z}+1} dz \\
&= 2\int\limits_{1}^{\mathrm{e}^{u}} \frac{1}{y^{2}+1} dy \\
&= 2\tan^{-1}\mathrm{e}^{u} \, -\frac{\pi}{2}
\tag{2}
\label{eq:g2}
\end{align}

## Derivative of the Gudermannian

\frac{d}{dx}\mathrm{gd}x = \mathrm{sech}(x)

follows from the integral definition.

## Maclaurin Series

$$\mathrm{gd}(0) = 0$$ follows from the integral definition.

\mathrm{gd}^{\prime}(0) = \mathrm{sech}(0) = 1

\mathrm{gd}^{\prime \prime}(0) = \mathrm{sech}^{\prime}(0) = -\mathrm{sech}(0)\tanh(0) = 0

\mathrm{gd}^{\prime \prime \prime}(0) = \mathrm{sech}^{\prime \prime}(0) = \mathrm{sech}(0)\tanh^{2}(0) – \mathrm{sech}^{3}(0) = -1

\mathrm{gd}x = \sum\limits_{n=0}^{\infty} \frac{x^{n}}{n!} \frac{d^{n}\mathrm{gd}(0)}{dx^{n}}
= x \, – \frac{x^{3}}{6} + \cdots

## The Gudermannian is an Odd Function

$$\mathrm{gd}(-x) = -\mathrm{gd}x$$ follows from the Maclaurin series.

## The Exponential Function and the Gudermannian

First:
\begin{align}
\mathrm{e}^{x} &= \sec(\mathrm{gd}x) + \tan(\mathrm{gd}x) \\
&= \cosh x + \sinh x \\
&= \frac{\mathrm{e}^{x} + \mathrm{e}^{-x}}{2} + \frac{\mathrm{e}^{x} – \mathrm{e}^{-x}}{2} \\
&= \mathrm{e}^{x}
\end{align}
Second:
\begin{align}
\mathrm{e}^{x} &= \frac{1+\sin(\mathrm{gd}x)}{\cos(\mathrm{gd}x)} \\
&= \sec(\mathrm{gd}x) + \tan(\mathrm{gd}x) \\
&= \mathrm{e}^{x}
\end{align}
Third:
\begin{align}
\mathrm{e}^{x} &= \tan\left(\frac{\pi}{4} + \frac{1}{2}\mathrm{gd}x \right) \\
&= \sec(\mathrm{gd}x) + \tan(\mathrm{gd}x) \\
&= \mathrm{e}^{x}
\end{align}

## Integral of the Gudermannian

Begin with Equation \eqref{eq:g2}, $$\mathrm{gd}x = 2\tan^{-1}\mathrm{e}^{x} \, – \pi/2$$ so that we have

\int \mathrm{gd}x \, dx = 2\int \tan^{-1}\mathrm{e}^{x} dx – \int \frac{\pi}{2} dx

Using

\tan^{-1}z = \frac{i}{2}\Big[\ln(1-iz) – \ln(1+iz)\Big]

and integrating this expression, yields
\begin{align}
\int \tan^{-1}\mathrm{e}^{x} dx &= \frac{i}{2}\int\Big[\ln(1-i\mathrm{e}^{x}) – \ln(1+i\mathrm{e}^{x})\Big] dx \\
&= \frac{i}{2} [-\mathrm{Li}_{2}(i\mathrm{e}^{x}) + \mathrm{Li}_{2}(-i\mathrm{e}^{x})]
\end{align}
We used the substitution $$y=-i\mathrm{e}^{x}$$ and the dilogarithm function $$\mathrm{Li}_{2}(z)$$.
Putting the pieces together, we obtain

\int \mathrm{gd}x \, dx = i[\mathrm{Li}_{2}(-i\mathrm{e}^{x})-\mathrm{Li}_{2}(i\mathrm{e}^{x})] – \frac{\pi}{2}x

## References

1. Wolfram MathWorld
2. Wikipedia
3. A Treatise on the Integral Calculus – Joseph Edwards; Chapter 3, Article 69.