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Integral challenge #1

DreamWeaver

Well-known member
Sep 16, 2013
337
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:


\(\displaystyle \int_0^{\infty}\frac{x^m}{a \sinh x +b \cosh x} \, dx =\)

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