## 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.

## 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.

## 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 $$\int \frac{1}{x^{n}(T-x)^{n}} dx$$

This question was posed at Mathematics Stack Exchange, here is my solution.

For $$n \in \mathbb{N}$$.

\begin{align}
\int \frac{1}{x^{n}(T-x)^{n}} dx &= \frac{1}{T^{n}} \int \frac{1}{x^{n}(1-x/T)^{n}} dx \\
&= T^{1-2n} \int y^{-n} (1-y)^{-n} dy \\
&= T^{1-2n} \mathrm{B}_{y}(1-n,1-n) \\
&= T^{1-2n} \mathrm{B}_{x/T}(1-n,1-n) \\
&= \frac{1}{1-n} \frac{x^{1-n}}{T^{n}} {}_{2}\mathrm{F}_{1}(1-n,n;2-n;x/T)
\end{align}

We have used the incomplete Beta function and Gauss’s hypergeometric function.

## Prove $$\lim_{n \to \infty} \int_{0}^{1} \frac{x^{n}}{\sqrt{1+x^{n}}} dx = 0$$

How to prove

\lim_{n \to \infty} \int\limits_{0}^{1} \frac{x^{n}}{\sqrt{1+x^{n}}} dx = 0

was a question posed at Mathematics Stack Exchange. There are more efficient answers there, but here is a fun and interesting solution.

Using the integral defintion (analytically continued) of Gauss’s hypergeometric function

{}_{2}\mathrm{F}_{1}(a,b;c;z) = \frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)} \int\limits_{0}^{1} t^{a-1} (1-t)^{c-b-1} (1-zt)^{-a} dt

for $$\mathrm{Re}\,c \gt \mathrm{Re}\,b \gt 0,\,\,|\mathrm{arg}(1-z)| \lt \pi$$

We have, using the substitution $$y=x^{n}$$
\begin{align}
\int\limits_{0}^{1} \frac{x^{n}}{\sqrt{1+x^{n}}} dx
&= \frac{1}{n} \int\limits_{0}^{1} \frac{y^{1/n}}{\sqrt{1+y}} dy \\
\tag{1}
&= \frac{\Gamma(1+1/n)\Gamma(1)}{n\Gamma(2+1/n)} \,{}_{2}\mathrm{F}_{1}\left(\frac{1}{2},1+\frac{1}{n};2+\frac{1}{n};-1 \right) \\
&= \frac{1}{n+1} \,{}_{2}\mathrm{F}_{1}\left(\frac{1}{2},1+\frac{1}{n};2+\frac{1}{n};-1 \right)
\end{align}

Taking the limit of the hypergeometric function and applying the integral definition again yields
\begin{align}
\lim_{n \to \infty} {}_{2}\mathrm{F}_{1}\left(\frac{1}{2},1+\frac{1}{n};2+\frac{1}{n};-1 \right)
&= {}_{2}\mathrm{F}_{1}\left(\frac{1}{2},1;2;-1 \right) \\
&= \frac{\Gamma(2)}{\Gamma(1)\Gamma(1)} \int\limits_{0}^{1} \frac{1}{\sqrt{1+t}} dt \\
&= 2(\sqrt{2} – 1)
\end{align}

Substituting this result into equation 1, we have

\lim_{n \to \infty} \int\limits_{0}^{1} \frac{x^{n}}{\sqrt{1+x^{n}}} dx = 0*2(\sqrt{2} – 1) = 0

## Evaluate $$\int_{0}^{1} \frac{x^{n}}{\sqrt{x^{3}+1}} dx$$

How to evaluate

\int\limits_{0}^{1} \frac{x^{n}}{\sqrt{x^{3}+1}} dx

was a question posed at Mathematics Stack Exchange. Here is my solution.

Let $$y=x^{3}$$
\begin{align}
\int\limits_{0}^{1} \frac{x^{n}}{\sqrt{x^{3}+1}} dx
&= \frac{1}{3} \int\limits_{0}^{1} \frac{y^{(n-2)/3}}{\sqrt{y+1}} dy \\
&= \frac{1}{3} \frac{\Gamma(\frac{n+1}{3})\Gamma(1)}{\Gamma(\frac{n+4}{3})}\,{}_{2}\mathrm{F}_{1}\left(\frac{1}{2},\frac{n+1}{3};\frac{n+4}{3};-1\right) \\
&= \frac{1}{n+1} \,{}_{2}\mathrm{F}_{1}\left(\frac{1}{2},\frac{n+1}{3};\frac{n+4}{3};-1\right)
\end{align}

We used the analytic continuation of Gauss’s hypergeometric function

{}_{2}\mathrm{F}_{1}(a,b;c;z)
= \frac{\Gamma(c)}{\Gamma(b)\Gamma(c – b)} \int\limits_{0}^{1} t^{b-1} (1-t)^{c-b-1} (1-zt)^{-a} dt

For $$\mathrm{Re}\, c \gt \mathrm{Re}\, b \gt 0 \, , \, |\mathrm{arg}(1-z)| \lt \pi$$

## Evaluate $$\int \frac{x^{2-2/n}}{(3-x)^{1/n}} dx$$

How to evaluate

\int \frac{x^{2-2/n}}{(3-x)^{1/n}} dx

was a question posed at Mathematics Stack Exchange. Here is my solution.

Let $$x=3z$$

\begin{align}
\int \frac{x^{2-2/n}}{(3-x)^{1/n}} dx &= 3^{-1/n} \int \frac{x^{2-2/n}}{(1-x/3)^{1/n}} dx \\
&= 3^{3-3/n} \int z^{2-2/n} (1-z)^{-1/n} dz \\
&= 3^{3-3/n} \mathrm{B}_{z} \left( 3-\frac{2}{n}, 1-\frac{1}{n} \right) \\
&= 3^{3-3/n} \frac{n}{3n-2} \, z^{3-2/n} \, {}_{2}\mathrm{F}_{1}\left(3-\frac{2}{n},\frac{1}{n};4-\frac{2}{n};z \right) \\
&= \frac{1}{3^{1/n}} \frac{n}{3n-2} \, x^{3-2/n} \, {}_{2}\mathrm{F}_{1}\left(3-\frac{2}{n},\frac{1}{n};4-\frac{2}{n};\frac{x}{3} \right)
\end{align}

Note:

\begin{align}
\mathrm{B}_{z}(p,q) &= \int_{0}^{z} t^{p-1} (1-t)^{q-1} \mathrm{d}t \\
&= \frac{z^{p}}{p} \,{}_{2}\mathrm{F}_{1}(p,1-q;p+1;z)
\end{align}
The incomplete beta function and hypergeometric function.

## Evaluate the Integral $$\int x\left(1+ax^{-k}\right)^{-m} \mathrm{d}x$$

How to evaluate

\int x\left(1+\frac{a}{x^{k}}\right)^{-m} \mathrm{d}x

was part of a question at Mathematics Stack Exchange. Here is my solution.

Let $$y = -\frac{a}{x^{k}}$$
\begin{align}
\int x\left(1+\frac{a}{x^{k}}\right)^{-m} \mathrm{d}x &=
\frac{1}{k} (-1)^{1+\frac{2}{k}} a^{\frac{2}{k}} \int (1-y)^{-m} y^{-1-\frac{2}{k}} \mathrm{d}y \\
&= \frac{1}{k} (-1)^{1+\frac{2}{k}} a^{\frac{2}{k}} \mathrm{B}_{y} \left(-\frac{2}{k},1-m \right) \\
&= \frac{1}{k} (-1)^{1+\frac{2}{k}} a^{\frac{2}{k}} y^{-\frac{2}{k}} (-1) \frac{k}{2}
\,{}_{2}\mathrm{F}_{1} \left(-\frac{2}{k},m;1-\frac{2}{k};y \right) \\
&= \frac{x^{2}}{2} \,{}_{2}\mathrm{F}_{1} \left(-\frac{2}{k},m;1-\frac{2}{k};-\frac{a}{x^{k}} \right)
\end{align}

Notes:
1.

\mathrm{B}_{z}(p,q) = \int_{0}^{z} t^{p-1} (1-t)^{q-1} \mathrm{d}t

is the incomplete beta function.
2.

\mathrm{B}_{z}(p,q) = \frac{z^{p}}{p} \,{}_{2}\mathrm{F}_{1}(p,1-q;p+1;z)

is the incomplete beta function in terms of Gauss’s hypergeometric function.

## Evaluate the Integral $$\int x^{km+1}(a+x^{k})^{-m} \mathrm{d}x$$

How to evaluate

\int x^{km+1}(a+x^{k})^{-m} \mathrm{d}x

was part of a question at Mathematics Stack Exchange. Here is my solution.

Let $$y = -\frac{x^{k}}{a}$$
\begin{align}
\int x^{km+1}(a+x^{k})^{-m} \mathrm{d}x &= \frac{1}{k}a^{2/k}(-1)^{m+2/k} \int (1-y)^{-m} y^{m-1+2/k} \mathrm{d}y \\
&= \frac{1}{k}a^{2/k}(-1)^{m+2/k} \mathrm{B}_{y}\left(m+\frac{2}{k},1-m\right) \\
&= \frac{1}{k}a^{2/k}(-1)^{m+2/k} \frac{y^{m+2/k}}{m+2/k} \,{}_{2}\mathrm{F}_{1}\left(m+\frac{2}{k},m;m+1+\frac{2}{k};y\right) \\
&= \frac{1}{km+2} \,\frac{1}{a^{m}} x^{km+2} \,{}_{2}\mathrm{F}_{1}\left(m+\frac{2}{k},m;m+1+\frac{2}{k};-\frac{x^{k}}{a}\right)
\end{align}

Notes:
1.

\mathrm{B}_{z}(p,q) = \int_{0}^{z} t^{p-1} (1-t)^{q-1} \mathrm{d}t

is the incomplete beta function.
2.

\mathrm{B}_{z}(p,q) = \frac{z^{p}}{p} \,{}_{2}\mathrm{F}_{1}(p,1-q;p+1;z)

is the incomplete beta function in terms of Gauss’s hypergeometric function.

## A new proof of the fundamental two-term transformation for the series $$_{3}\mathrm{F}_{2}(1)$$ due to Thomae by Rathie

The aim of this short note is to provide a very simple proof for obtaining the fundamental two-term transformation for the series $$_{3}\mathrm{F}_{2}(1)$$ due to Thomae.

The paper can be found here.