## Integrate $$\int_{0}^{\infty} \frac{x}{\sqrt{\mathrm{e}^{x}-1}} \mathrm{d} x$$

How to evaluate

\int\limits_{0}^{\infty} \frac{x}{\sqrt{\mathrm{e}^{x}-1}} \mathrm{d} x

was a question on Mathematics Stack Exchange. Here is another solution method.

Let $$z=\mathrm{e}^{x}-1$$, so that we have

\int\limits_{0}^{\infty} \frac{\mathrm{ln}(z+1)}{\sqrt{z}}\frac{1}{z+1} \mathrm{d} z

Let us consider

I(a) = \int\limits_{0}^{\infty} \frac{(z+1)^{a}}{\sqrt{z}} \mathrm{d} z = \mathrm{B}\left(\frac{1}{2}, -\frac{1}{2}-a\right)
= \frac{\Gamma\left(\frac{1}{2}\right) \Gamma\left(-\frac{1}{2}-a\right)}{\Gamma(-a)}

so that

\lim_{a \to -1} \frac{\partial I(a)}{\partial a} = \int\limits_{0}^{\infty} \frac{\mathrm{ln}(z+1)}{\sqrt{z}}\frac{1}{z+1} \mathrm{d} z =
\int\limits_{0}^{\infty} \frac{x}{\sqrt{\mathrm{e}^{x}-1}} \mathrm{d} x

Then,

\frac{\partial I(a)}{\partial a} = \Gamma\left(\frac{1}{2}\right)\left[\frac{-\Gamma(-a)\Gamma\left(-\frac{1}{2}-a\right)\psi^{0}\left(-\frac{1}{2}-a\right) + \Gamma\left(-\frac{1}{2}-a\right)\Gamma(-a)\psi^{0}(-a)}{\Gamma(-a)\Gamma(-a)} \right]

\begin{align}
\lim_{a \to -1} \frac{\partial I(a)}{\partial a} & = \frac{-\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma(1)}
\left[\psi^{0}\left(\frac{1}{2}\right) – \psi^{0}(1)\right] \\
& = -\pi[(-\gamma-\mathrm{ln}4) -(- \gamma)] \\
& = \pi\mathrm{ln}4 \\
& = \int\limits_{0}^{\infty} \frac{x}{\sqrt{\mathrm{e}^{x}-1}} \mathrm{d} x
\end{align}

## Integrate $$\int_{0}^{\infty} \frac{\mathrm{ln}^{3}(x)}{(1+x^{2})(1+x)^{2}} \mathrm{d} x$$

The following integral was a question on Mathematics Stack Exchange.

\int\limits_{0}^{\infty} \frac{\mathrm{ln}^{3}(x)}{(1+x^{2})(1+x)^{2}} \mathrm{d} x
\label{eq:160813a1}
\tag{1}

As usual, there are multiple clever solutions. However, a user noted that the integral could be evaluated via the Mellin transform but he did not provide any details so I will do it here.

Let us evaluate it via the Mellin transform

\mathcal{M}[f(x)](s) = \int\limits_{0}^{\infty} x^{s-1} f(x) \mathrm{d} x
\label{eq:160813a2}
\tag{2}

where

f(x) = \frac{1}{(1+x^{2})(1+x)^{2}} = -\frac{1}{2}\frac{x}{x^{2}+1} + \frac{1}{2}\frac{1}{x+1} + \frac{1}{2}\frac{1}{(x+1)^{2}}
\label{eq:160813a3}
\tag{3}

via partial fraction expansion.

Applying the Mellin transform, yields
\begin{align}
\mathcal{M}[f(x)](s) & = \int\limits_{0}^{\infty} \frac{x^{s-1}}{(1+x^{2})(1+x)^{2}} \\
& = -\frac{1}{2}\left[\frac{1}{2}\pi\sec\left(\frac{\pi}{2}s\right)\right] + \frac{1}{2}\pi\csc(\pi s) + \frac{1}{2} \mathrm{B}(s,2-s)
\label{eq:160813a4}
\tag{4}
\end{align}

Taking the 3rd derivative of equation \eqref{eq:160813a4} with respect to s and then taking $$\lim s \to 1$$ yields
\begin{align}
\int\limits_{0}^{\infty} \frac{\mathrm{ln}^{3}(x)}{(1+x^{2})(1+x)^{2}} \mathrm{d} x & = -\frac{1}{2}\left( -\frac{7}{960} \pi^{4} \right) + \frac{1}{2} \left( -\frac{7}{60} \pi^{4} \right) + \frac{1}{2}0 \\
& = -\frac{7}{128} \pi^{4}
\label{eq:160813a5}
\tag{5}
\end{align}

Let us fill in the details. Handling the beta function first, we have

\mathrm{B}(s,2-s) = \frac{\Gamma(s)\Gamma(2-s)}{\Gamma(2)} = \Gamma(s)\Gamma(2-s)
\label{eq:160813a6}
\tag{6}

To take derivatives, we note that

\frac{\mathrm d}{\mathrm d s} \Gamma(s) = \Gamma(s) \psi^{(0)}(s) \quad \mathrm{and} \quad \psi^{(n)}(s) = \frac{\mathrm{d}^{n}}{\mathrm{d} s^{n}} \psi^{(0)}(s)

Where $$\psi^{(n)}(s)$$ is the polygamma function.

Taking the third derivative of equation \eqref{eq:160813a6} and letting $$\lim s \to 1$$ equals 0. Here we used

and fortunately $$\Gamma^{(3)}(s) = -\Gamma^{(3)}(2-s)$$ which leads to some cancellations.

Doing the same for the first two terms on the right hand side of equation \eqref{eq:160813a4}, we have to be careful. Each of them individually goes to $$\infty$$ as $$\lim s \to 1$$ but the $$(s-1)^{-4}$$ terms in the Laurent expansions about $$s=1$$ cancel. Here I used Wolfram Alpha. Doing it with the two terms combined yielded our final answer.

## The nth Catalan Number Expressed as a Beta Function

John Cook who blogs here, posted the following expression for the nth Catalan number on one of his twitter accounts

C_{n} = \frac{1}{2\pi} \int\limits_{0}^{4} x^{n} \sqrt{\frac{4-x}{x}} \mathrm{d} x
\label{eq:1608121}
\tag{1}

Let us express this as a Beta function
\begin{align}
\int\limits_{0}^{4} x^{n} \sqrt{\frac{4-x}{x}} \mathrm{d} x & = x^{n-\frac{1}{2}} (4-x)^{\frac{1}{2}} \mathrm{d} x \\
& = 4^{n+1} \int\limits_{0}^{1} y^{n-\frac{1}{2}} (1-y)^{\frac{1}{2}} \mathrm{d} y \\
& = 4^{n+1} \, \mathrm{B}\left(n + \frac{1}{2}, \frac{3}{2} \right)
\label{eq:1608122}
\tag{2}
\end{align}
We used the substitution $$y=\frac{x}{4}$$.

Now we have

C_{n} = \frac{1}{2\pi} \int\limits_{0}^{4} x^{n} \sqrt{\frac{4-x}{x}} \mathrm{d} x = \frac{2^{2n+1}}{\pi} \mathrm{B}\left(n + \frac{1}{2}, \frac{3}{2} \right)
\label{eq:1608123}
\tag{3}

We will check this by using the following definition of the nth Catalan number

C_{n} = \frac{(2n)!}{(n+1)! \, n!}
\label{eq:1608124}
\tag{4}

Using the following Gamma function expressions

\Gamma\left(n + \frac{1}{2}\right) = \frac{(2n)!\sqrt{\pi}}{4^{n}n!}

We have
\begin{align}
\frac{2^{2n+1}}{\pi} \mathrm{B}\left(n + \frac{1}{2}, \frac{3}{2} \right) & = \frac{2^{2n+1}}{\pi} \frac{\Gamma\left(n + \frac{1}{2}\right) \Gamma\left(\frac{3}{2}\right)}{\Gamma(n+2)} \\
& = \frac{2^{2n+1}}{\pi} \frac{(2n)!\sqrt{\pi}}{2^{2n}n!} \frac{\sqrt{\pi}}{2} \frac{1}{(n+1)!} \\
& = \frac{(2n)!}{(n+1)! \, n!}
\label{eq:1608125}
\tag{5}
\end{align}
thus equation \eqref{eq:1608123} is correct.

## Derivation of Beta Function Expressions

In this post, I will derive some basic expressions of the beta function. I will follow the equation numbering of Higher Transcendental Functions (Bateman Manuscript), Volume 1, page 9 (print), page 35 (pdf).

We begin with the basic integral definition of the beta function

\mathrm{B}(x,y) = \int\limits^{1}_{0} t^{x-1} (1-t)^{y-1} \mathrm{d} t
\quad \mathrm{for} \,\, \mathrm{Re} \, x > 0 \,\, \mathrm{and} \,\, \mathrm{Re} \, y > 0
\label{eq:bf1}
\tag{1}

\mathrm{B}(x,y) = \int\limits_{0}^{\infty} \frac{v^{x-1}}{(1+v)^{x+y}} \mathrm{d} v
\quad \mathrm{for} \,\, \mathrm{Re} \, x > 0 \,\, \mathrm{and} \,\, \mathrm{Re} \, y > 0
\label{eq:bf2}
\tag{2}

To derive this equation, we begin with \eqref{eq:bf1} and make the substitution

t = \frac{v}{1+v}

to obtain

\mathrm{B}(x,y) = \int\limits_{0}^{\infty} \frac{v^{x-1}}{(1+v)^{x-1}} \Big(1 – \frac{v}{1+v}\Big)^{y-1} \frac{1}{(1+v)^{2}} \mathrm{d} v

simplification yields \eqref{eq:bf2}.

\mathrm{B}(x,y) = \int\limits_{0}^{1} \frac{(v^{x-1}+v^{y-1})}{(1+v)^{x+y}} \mathrm{d} v
\quad \mathrm{for} \,\, \mathrm{Re} \, x > 0 \,\, \mathrm{and} \,\, \mathrm{Re} \, y > 0
\label{eq:bf3}
\tag{3}

To obtain \eqref{eq:bf3} we begin with \eqref{eq:bf2} and break up the integral

\mathrm{B}(x,y) = \int\limits_{0}^{1} \frac{v^{x-1}}{(1+v)^{x+y}} \mathrm{d} v + \int\limits_{1}^{\infty} \frac{v^{x-1}}{(1+v)^{x+y}} \mathrm{d} v

and designate the last integral as I.

For I, we make the substitution $$w = v^{-1}$$

\mathrm{I} = \int\limits_{0}^{1} \frac{1}{w^{x-1}} \frac{w^{x+y}}{(1+w)^{x+y}} \frac{1}{w^{2}} \mathrm{d} w

Simplifying and then making the substitution for I yields \eqref{eq:bf3}.

\mathrm{B}(x,y) = \mathrm{B}(y,x)
\label{eq:bf4}
\tag{4}

Equation \eqref{eq:bf4} follows directly from equation \eqref{eq:bf3}.

\mathrm{B}(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)}
\label{eq:bf5}
\tag{5}

To derive equation \eqref{eq:bf5}, let us start with the basic integral definition of the gamma function and then make the substitution $$x = at$$

\begin{align}
\Gamma(z) & = \int\limits_{0}^{\infty} x^{z-1} \mathrm{e}^{-x} \mathrm{d} x \\
& = a^{z} \int\limits_{0}^{\infty} t^{z-1} \mathrm{e}^{-at} \mathrm{d} t \\
\end{align}

Rearranging terms yields

\frac{1}{a^{z}} = \frac{1}{\Gamma(z)} \int\limits_{0}^{\infty} t^{z-1} \mathrm{e}^{-at} \mathrm{d} t

and making the substitutions $$z = \alpha + \beta$$ and $$a = 1 + v$$, we obtain

\frac{1}{(1+v)^{\alpha + \beta}} = \frac{1}{\Gamma(\alpha + \beta)} \int\limits_{0}^{\infty} t^{\alpha + \beta – 1} \mathrm{e}^{-(1+v)t} \mathrm{d} t

Combining this with equation \eqref{eq:bf2}, we have

\begin{align}
\mathrm{B}(\alpha,\beta) & = \frac{1}{\Gamma(\alpha + \beta)} \int\limits_{0}^{\infty} \int\limits_{0}^{\infty} v^{\beta – 1} t^{\alpha + \beta – 1}
\mathrm{e}^{-t} \mathrm{e}^{-vt} \mathrm{d} t \mathrm{d} v \\
& = \frac{1}{\Gamma(\alpha + \beta)} \int\limits_{0}^{\infty} \Big[\int\limits_{0}^{\infty} t^{\alpha – 1} \mathrm{e}^{-t} \mathrm{d} t \Big] t^{\beta} v^{\beta – 1} \mathrm{e}^{-vt} \mathrm{d} v \\
& = \frac{\Gamma(\alpha)}{\Gamma(\alpha + \beta)} \int\limits_{0}^{\infty} t^{\beta} v^{\beta – 1} \mathrm{e}^{-vt} \mathrm{d} v \\
& = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha + \beta)} \\
\end{align}

\mathrm{B}(x,y+1) = \frac{y}{x}\mathrm{B}(x+1,y) = \frac{y}{x+y}\mathrm{B}(x,y)
\label{eq:bf6}
\tag{6}

To derive equation \eqref{eq:bf6}, use \eqref{eq:bf5} to convert the beta functions into gamma functions and use

\Gamma(z+1) = z\Gamma(z)

\begin{align}
\mathrm{B}(x,y+1) & = \frac{\Gamma(x)\Gamma(y+1)}{\Gamma(x+y+1)} \\
& = \frac{y\Gamma(x)\Gamma(y)}{(x+y)\Gamma(x+y)} \\
& = \frac{y\mathrm{B}(x,y)}{x+y} \\
\end{align}

\begin{align}
\mathrm{B}(x+1,y) & = \frac{\Gamma(x+1)\Gamma(y)}{\Gamma(x+y+1)} \\
& = \frac{x\Gamma(x)\Gamma(y)}{(x+y)\Gamma(x+y)} \\
& = \frac{y\mathrm{B}(x,y)}{x+y} \\
\end{align}

\mathrm{B}(x,y)\mathrm{B}(x+y,z) = \mathrm{B}(y,z)\mathrm{B}(y+z,x) = \mathrm{B}(z,x)\mathrm{B}(x+z,y)
\label{eq:bf7}
\tag{7}

For equation \eqref{eq:bf7} we convert the beta functions to gamma functions.
\begin{align}
\mathrm{B}(x,y)\mathrm{B}(x+y,z) & = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)} \times \frac{\Gamma(x + y)\Gamma(z)}{\Gamma(x + y + z)} \times
\frac{\Gamma(y + z)}{\Gamma(y + z)} \\
& = \frac{\Gamma(y)\Gamma(z)}{\Gamma(y + z)} \times \frac{\Gamma(y + z)\Gamma(x)}{\Gamma(x + y + z)} \\
& = \mathrm{B}(y,z)\mathrm{B}(y+z,x) \\
\end{align}

\begin{align}
\mathrm{B}(x,y)\mathrm{B}(x+y,z) & = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y)} \times \frac{\Gamma(x + y)\Gamma(z)}{\Gamma(x + y + z)} \times
\frac{\Gamma(x + z)}{\Gamma(x + z)} \\
& = \frac{\Gamma(z)\Gamma(x)}{\Gamma(z + x)} \times \frac{\Gamma(x + z)\Gamma(y)}{\Gamma(x + y + z)} \\
& = \mathrm{B}(z,x)\mathrm{B}(x+z,y) \\
\end{align}

\mathrm{B}(x,y)\mathrm{B}(x+y,z)\mathrm{B}(x+y+z,u) = \frac{\Gamma(x)\Gamma(y)\Gamma(z)\Gamma(u)}{\Gamma(x + y + z + u)}
\label{eq:bf8}
\tag{8}

Here, we use equation \eqref{eq:bf5}

\mathrm{B}(x,y)\mathrm{B}(x+y,z)\mathrm{B}(x+y+z,u) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x + y} \times \frac{\Gamma(x+y)\Gamma(z)}{\Gamma(x + y + z} \times \frac{\Gamma(x+y+z)\Gamma(u)}{\Gamma(x + y + z + u}

The generalization of equation \eqref{eq:bf8} is evident.

\frac{1}{\mathrm{B}(n,m)} = m \binom{n+m-1}{n-1} = n \binom{n+m-1}{m-1} \,\, \mathrm{for} \,\, n,m \in \mathbb{Z}^{+}
\label{eq:bf9}
\tag{9}

We use the following relationships and then invoke equation \eqref{eq:bf5}

\frac{1}{\mathrm{B}(n,m)} = \frac{\Gamma(n + m)}{\Gamma(n)\Gamma(m)} = \frac{(n + m – 1)!}{(n – 1)!(m – 1)!} =
m\frac{(n + m – 1)!}{m!(n – 1)!} = m \binom{n+m-1}{n-1}

We can do the same for the last part of equation \eqref{eq:bf9}.

Additional information about the beta function can be found at the following online references

## A Laplace Transform Proof of $$\mathrm{B}(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$

This proof appeared in Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals by George Boros and Victor Moll. Here, I fill in the steps of the proof.

We begin with 3 fundamental results of the Laplace transform: the basic definition, convolution, and the Laplace transform of a convolution.

\mathcal{L}[f(t)](s) = \int\limits_{0}^{\infty} \mathrm{e}^{-st} f(t) \mathrm{d} t
\label{eq:1608061}
\tag{1}

(f*g)(t) = \int\limits_{0}^{t} f(\tau) g(t-\tau) \mathrm{d} \tau
\label{eq:1608062}
\tag{2}

\mathcal{L}[f*g] = \mathcal{L}[f]\mathcal{L}[g]
\label{eq:1608063}
\tag{3}

We begin with the following two functions

\label{eq:1608064}
\tag{4}

These functions were chosen for two reasons. First, they appear in the basic integral definition of the beta function

\mathrm{B}(x,y) = \int\limits_{0}^{1} t^{x-1} (1-t)^{y-1} \mathrm{d} t
\label{eq:1608065}
\tag{5}

and, as we now show, their Laplace transforms result in gamma functions
\begin{align}
\mathcal{L}[f(t)] & = \mathcal{L}[t^{x-1}] = \int\limits_{0}^{\infty} \mathrm{e}^{-st} t^{x-1} \mathrm{d} t \\
& = \frac{1}{s^{x}} \int\limits_{0}^{\infty} \mathrm{e}^{-v} v^{x-1} \mathrm{d} v \\
& = \frac{\Gamma(x)}{s^{x}}
\label{eq:1608066}
\tag{6}
\end{align}
where we have used the substitution $$v = st$$. Likewise, taking the Laplace transform of $$g(t)$$ yields,

\mathcal{L}[g(t)] = \mathcal{L}[t^{y-1}] = \frac{\Gamma(y)}{s^{y}}
\label{eq:1608067}
\tag{7}

Now we substitute equations \eqref{eq:1608067}, \eqref{eq:1608066}, and \eqref{eq:1608062} into \eqref{eq:1608063}

\frac{\Gamma(x)}{s^{x}} \frac{\Gamma(y)}{s^{y}} = \mathcal{L}\left[\int\limits_{0}^{t} \tau^{x-1} (t – \tau)^{y-1} \mathrm{d} \tau \right]
\label{eq:1608068}
\tag{8}

To evaluate the integral, we let $$\tau = tu$$
\begin{align}
\int\limits_{0}^{t} \tau^{x-1} (t – \tau)^{y-1} \mathrm{d} \tau & = t^{x+y-1} \int\limits_{0}^{1} u^{x-1} (1 – u)^{y-1} \mathrm{d} u \\
& = t^{x+y-1} \mathrm{B}(x,y)
\label{eq:1608069}
\tag{9}
\end{align}

Thus, we have
\begin{align}
\mathcal{L}[t^{x+y-1} \mathrm{B}(x,y)] & = \frac{\Gamma(x)}{s^{x}} \frac{\Gamma(y)}{s^{y}} \\
& = \mathrm{B}(x,y) \int\limits_{0}^{\infty} \mathrm{e}^{-st} t^{x+y-1} \mathrm{d} t \\
& = \mathrm{B}(x,y) \frac{1}{s^{x+y}} \int\limits_{0}^{\infty} \mathrm{e}^{-v} v^{x+y-1} \mathrm{d} v \\
& = \mathrm{B}(x,y) \frac{\Gamma(x+y)}{s^{x+y}}
\label{eq:16080610}
\tag{10}
\end{align}

Combining the first and last results from the right hand side of the above equation yields our final result

\mathrm{B}(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}

## A Derivation of the Beta Function Representation $$\mathrm{B}(x,y) = 2^{1-x-y} \int_{0}^{1} (1+t)^{x-1}(1-t)^{y-1} + (1+t)^{y-1}(1-t)^{x-1} \mathrm{d} t$$

I used the beta function representation

\mathrm{B}(x,y) = 2^{1-x-y} \int\limits_{0}^{1} (1+t)^{x-1}(1-t)^{y-1} + (1+t)^{y-1}(1-t)^{x-1} \mathrm{d} t
\label{eq:1608051}
\tag{1}

here to evaluate an integral. Now I will derive this result.

We begin with the basic integral definition of the beta function

\mathrm{B}(x,y) = \int\limits^{1}_{0} t^{x-1} (1-t)^{y-1} \mathrm{d} t
\label{eq:1608052}
\tag{2}

then use the substitution $$t = \frac{z-a}{b-a}$$, this yields

\begin{align}
\mathrm{B}(x,y) & = \frac{1}{b-a} \int\limits^{b}_{a} \left(\frac{z-a}{b-a}\right)^{x-1} \left(\frac{b-z}{b-a}\right)^{y-1} \mathrm{d} z \\
& = (b-a)^{-x-y+1} \int\limits^{b}_{a} (z-a)^{x-1} (b-z)^{y-1} \mathrm{d} z \\
& = 2^{-x-y+1} \int\limits_{-1}^{1} (z+1)^{x-1} (1-z)^{y-1} \mathrm{d} z \\
& = 2^{-x-y+1} \left[ \int\limits_{-1}^{0} (z+1)^{x-1} (1-z)^{y-1} \mathrm{d} z + \int\limits_{0}^{1} (z+1)^{x-1} (1-z)^{y-1} \mathrm{d} z \right]
\label{eq:1608053}
\tag{3}
\end{align}

Note that we let $$a=-1$$ and $$b=1$$.

For the rightmost integral, we let $$z=-w$$ so that

\int\limits_{0}^{1} (z+1)^{x-1} (1-z)^{y-1} \mathrm{d} z = \int\limits_{0}^{1} (1-w)^{x-1} (1+w)^{y-1} \mathrm{d} w
\label{eq:1608054}
\tag{4}

Substituting equation \eqref{eq:1608054} into equation \eqref{eq:1608053} yields equation \eqref{eq:1608051}.

The original substitution, $$t = \frac{z-a}{b-a}$$ is a useful method of transforming the limits of integration from 0 to 1 into a to b.

## Integrate $$\int_{0}^{a} (a^{2}-x^{2})^{n-1/2} \mathrm{d} x$$

From Victor Moll’s attempt to solve all of the integrals in Gradshteyn and Ryzhik, we have

\int\limits_{0}^{a} (a^{2}-x^{2})^{n-1/2} \mathrm{d} x = a^{2n} \frac{(2n-1)!!}{(2n)!!} \frac{\pi}{2}
\label{eq:1608041}
\tag{1}

Moll’s solution is at the link, so let us solve this another way. We first use the substitution used by Moll, let $$x=av$$ to obtain

\int\limits_{0}^{a} (a^{2}-x^{2})^{n-1/2} \mathrm{d} x = a^{2n} \int\limits_{0}^{1} (1-v^{2})^{n-1/2} \mathrm{d} v
\label{eq:1608042}
\tag{2}

We designate the integral on the right hand side as I and make the substitution $$z = v^{2}$$

\mathrm{I} = \int\limits_{0}^{1} (1-v^{2})^{n-1/2} \mathrm{d} v = \frac{1}{2} \int\limits_{0}^{1} z^{-1/2} (1-z)^{n-1/2} \mathrm{d} z
\label{eq:1608043}
\tag{3}

From the beginning this integral looked like a gamma or beta function, now, in this form it is obviously a beta function. We then have

\mathrm{I} = \frac{1}{2} \mathrm{B}\left(\frac{1}{2}, n + \frac{1}{2} \right)
\label{eq:1608044}
\tag{4}

and thus

\int\limits_{0}^{a} (a^{2}-x^{2})^{n-1/2} \mathrm{d} x = a^{2n} \frac{1}{2} \mathrm{B}\left(\frac{1}{2}, n + \frac{1}{2} \right)
\label{eq:1608045}
\tag{5}

Now we must show that the right hand sides of equations \eqref{eq:1608041} and \eqref{eq:1608045} are equal or

\mathrm{B}\left(\frac{1}{2}, n + \frac{1}{2} \right) \overset{\underset{\mathrm{?}}{}}{=} \pi \frac{(2n-1)!!}{(2n)!!}
\label{eq:1608046}
\tag{6}

We begin by expressing the beta function in terms of gamma functions

\mathrm{B}\left(\frac{1}{2}, n + \frac{1}{2} \right) = \frac{\Gamma(\frac{1}{2}) \Gamma(n + \frac{1}{2})}{\Gamma(n+1)} = \sqrt{\pi}
\frac{\Gamma(n + \frac{1}{2})}{\Gamma(n+1)}
\label{eq:1608047}
\tag{7}

Now we invoke the following two relationships between the gamma function and the double factorial, found at Wikipedia and Wolfram Math World respectively

(2n-1)!! = \frac{2^{n} \Gamma(n + \frac{1}{2})}{\sqrt{\pi}} \quad \mathrm{and} \quad (2n)!! = 2^{n}n! = 2^{n} \Gamma(n+1)

Substituting these expressions into \eqref{eq:1608046} shows that it is indeed an equality.

Putting everything together, we have

\int\limits_{0}^{a} (a^{2}-x^{2})^{n-1/2} \mathrm{d} x = a^{2n} \frac{(2n-1)!!}{(2n)!!} \frac{\pi}{2} = a^{2n} \frac{1}{2} \mathrm{B}\left(\frac{1}{2}, n + \frac{1}{2} \right)

## Integrate $$\int_{-1}^{1}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x$$

This integral appeared in Inside Interesting Integrals by Paul Nahin in the problem set of chapter 3. Using Wolfram Alpha, we get

\int\limits_{-1}^{1}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x = \pi
\label{eq:1}
\tag{1}

Nahin suggests the following trig substitution, $$x = \cos(2y)$$.

While the form of the integrand certainly does suggest that some type of trig substitution will work, let us do it with another method. If we write the integral as

\int\limits_{-1}^{1} (1+x)^{\frac{1}{2}}(1-x)^{-\frac{1}{2}} \mathrm{d} x

this looks like a beta function. From Higher Transcendental Functions (Bateman Manuscript), Volume 1, Section 1.5.1, equation 10, we see

\mathrm{B}(x,y) = 2^{1-x-y} \int\limits_{0}^{1} (1+t)^{x-1}(1-t)^{y-1} + (1+t)^{y-1}(1-t)^{x-1} \mathrm{d} t
\label{eq:2}
\tag{2}

Let us begin with the original integral and the right half of the interval of integration

\int\limits_{0}^{1} (1+x)^{\frac{1}{2}}(1-x)^{-\frac{1}{2}} \mathrm{d} x = \int\limits_{0}^{1}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x
\label{eq:3}
\tag{3}

Now, let us consider

\int\limits_{0}^{1} (1+x)^{-\frac{1}{2}}(1-x)^{\frac{1}{2}} \mathrm{d} x = \int\limits_{0}^{1}\sqrt{\frac{1-x}{1+x}} \mathrm{d} x
\label{eq:4}
\tag{4}

We let $$x=-y$$ to obtain

-\int\limits_{0}^{-1} \sqrt{\frac{1+y}{1-y}} \mathrm{d} y,
\label{eq:5}
\tag{5}

which we can rewrite as

\int\limits_{-1}^{0}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x
\label{eq:6}
\tag{6}

Adding the right hand side of equation \eqref{eq:3} and equation \eqref{eq:6} yields our original integral

\int\limits_{-1}^{0}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x + \int\limits_{0}^{1}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x = \int\limits_{-1}^{1}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x
\label{eq:7}
\tag{7}

Likewise, adding the left hand sides of equations \eqref{eq:4} and \eqref{eq:3} yields

\int\limits_{-1}^{0}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x + \int\limits_{0}^{1}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x =
\int\limits_{0}^{1} (1+x)^{-\frac{1}{2}}(1-x)^{\frac{1}{2}} \mathrm{d} x + \int\limits_{0}^{1} (1+x)^{\frac{1}{2}}(1-x)^{-\frac{1}{2}} \mathrm{d} x

If we combine this result into one integral and rearrange the integrand, we see that it is the same as the integral in \eqref{eq:2} with

x=\frac{3}{2} \,\, \mathrm{and} \,\, y=\frac{1}{2}

Putting it all together, we have

\int\limits_{-1}^{1}\sqrt{\frac{1+x}{1-x}} \mathrm{d} x = 2\mathrm{B}\left(\frac{3}{2},\frac{1}{2}\right) = \pi