Integral Challenge #1: Define & Prove Special Functions

[DreamWeaver]

Define the special functions:\(\displaystyle \text{Ti}_1(z)=\tan^{-1}z\)

\(\displaystyle \text{Ti}_{m+1}(z)=\int_0^z\frac{ \text{Ti}_{m+1}(x)}{x}\,dx\)and\(\displaystyle \text{Thi}_1(z)=\tanh^{-1}z\)

\(\displaystyle \text{Thi}_{m+1}(z)=\int_0^z\frac{ \text{Thi}_{m+1}(x)}{x}\,dx\)
Now, for \(\displaystyle a, b \in \mathbb{R}^{+}\), prove the following:\(\displaystyle \int_0^{\infty}\frac{x^m}{a \sinh x +b \cosh x} \, dx =\)

\(\displaystyle
\begin{cases}
2\frac{m! }{\sqrt{b^2-a^2}}\text{Ti}_{m+1} \left( \sqrt{\frac{b-a}{b+a}} \right), & \text{if } b>a>0 \\
2\frac{m! }{\sqrt{a^2-b^2}}\text{Thi}_{m+1} \left( \sqrt{\frac{a-b}{a+b}} \right), & \text{if } a>b>0
\end{cases}\)

 

[jvicens]

The special functions \text{Ti}_m(z) and \text{Thi}_m(z) are known as the inverse tangent and inverse hyperbolic tangent functions, respectively. They are defined as the inverse functions of the tangent and hyperbolic tangent functions, and are commonly used in mathematics and physics.

The first function, \text{Ti}_1(z), is simply the inverse tangent function, also denoted as \tan^{-1}z. It represents the angle whose tangent is equal to z. The second function, \text{Ti}_{m+1}(z), is a special case of the inverse tangent function where the angle is integrated over the interval [0,z].

Similarly, the first function, \text{Thi}_1(z), is the inverse hyperbolic tangent function, also denoted as \tanh^{-1}z. It represents the argument whose hyperbolic tangent is equal to z. The second function, \text{Thi}_{m+1}(z), is a special case of the inverse hyperbolic tangent function where the argument is integrated over the interval [0,z].

Now, to prove the given statement, we will use the properties of the special functions and the substitution method.

First, let us consider the case where b>a>0. We will substitute x = \sqrt{\frac{b-a}{b+a}}t in the given integral. This substitution will change the limits of integration to [0,\infty) and the integrand will become \frac{1}{\sqrt{b^2-a^2}} \frac{t^m}{\sin t}. Using the definition of the inverse tangent function, we can rewrite this as \frac{1}{\sqrt{b^2-a^2}} \text{Ti}_m(t). Therefore, the given integral becomes:

\begin{align*}
\int_0^{\infty}\frac{x^m}{a \sinh x +b \cosh x} \, dx &= \int_0^{\infty} \frac{1}{\sqrt{b^2-a^2}} \text{Ti}_m(t)\,dt \\
&= \frac{1}{\sqrt{b^2-a^2}} \int_0^{\infty} \text{Ti}_m(t)\,dt \\
&= \frac{1}{\sqrt{b^2