## Integrals of Gradshteyn and Ryzhik: 3.267

We use the following definitions of the gamma and beta functions:

\Gamma(z) = \int\limits_{0}^{\infty} \mathrm{e}^{-t} t^{z-1} dt, \,\, \mathfrak{R}(z) \gt 0

\mathrm{B}(a,b) = \int\limits_{0}^{1} t^{a-1} (1-t)^{b-1} dt
= \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}, \,\, \mathfrak{R}(a) \gt 0 \,\, \mathrm{and} \,\, \mathfrak{R}(b) \gt 0

## 3.267.1

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

## 3.267.2

\begin{align}
\int\limits_{0}^{1} x^{3n-1} (1-x^3)^{-1/3} dx
&= \frac{1}{3} \int\limits_{0}^{1} y^{n-1} (1-y)^{-1/3} dy
= \frac{1}{3} \mathrm{B}\left(n, \frac{2}{3} \right) \\
&= \frac{\Gamma(n)\Gamma(2/3)}{3\Gamma\left(n + \frac{2}{3}\right)}
= \frac{(n-1)! \Gamma(2/3)}{3\Gamma\left(n + \frac{2}{3}\right)}
\end{align}

## 3.267.3

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

we used the substitution $$y=x^3$$ for all three integrals.

## 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.

## For What Values of $$p$$ does $$\int_{1}^{\infty} \frac{1}{x^{2} \sqrt{x^{p}-1}} dx$$ Converge?

This was a question posed at Mathematics Stack Exchange. The obvious solutions were posted before I saw the question so this is an alternative solution that is interesting yet inefficient.

Let $$y=1/x$$, then $$z=y^{-p}-1$$
\begin{align}
\int\limits_{1}^{\infty} \frac{1}{x^{2} \sqrt{x^{p}-1}} dx &= \int\limits_{0}^{1} \frac{1}{\sqrt{y^{-p}-1}} dy \\
&= \frac{1}{p} \int\limits_{0}^{\infty} \frac{z^{-1/2}}{(z+1)^{1+1/p}} dz \\
&= \frac{1}{p} \mathrm{B}\left(\frac{1}{2},\frac{1}{p} + \frac{1}{2} \right) \\
&= \frac{\sqrt{\pi}\Gamma \left( \frac{1}{p} + \frac{1}{2} \right)}{\Gamma \left( \frac{1}{p} \right)}
\end{align}

The integrand and the limits of integration require that $$p \gt 0$$. From the
beta function we have $$\mathrm{Re} (\frac{1}{p} + \frac{1}{2}) \gt 0$$, $$p \lt -2$$ and $$p \gt 0$$. Thus we
must have $$p \gt 0$$.

We have used the following integral definition of the beta function

\mathrm{B}(x+1,y+1) = \int\limits_{0}^{\infty} \frac{z^{x}}{(1+z)^{x+y+2}} dz

## Prove $$\int_{0}^{1} (1-x^{1/k})^{n} dx = {1\over {k+n\choose n}}$$

How to prove

\int\limits_{0}^{1} (1-x^{1/k})^{n} dx = {1\over {k+n\choose n}}

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

Let $$y=x^{1/k}$$
\begin{align}
\int\limits_{0}^{1} (1-x^{1/k})^{n} dx &= k\int\limits_{0}^{1} (1-y)^{n} y^{k-1} dy \\
&= k\, \mathrm{B}(k,n+1) \\
&= \frac{k\,\Gamma(k)\Gamma(n+1)}{\Gamma(k+n+1)} \\
&= \frac{k!n!}{(k+n)!} \\
&= {1\over {k+n\choose n}}
\end{align}

## 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.

## Integrate $$\int_{0}^{\infty} \frac{x^{a}}{1+x^{2}} \mathrm{d}x$$

How to evaluate this integral was a question at Mathematics Stack Exchange. The first method we present was already answered at MSE but here we fill in the missing steps.

## Method 1

Let $$z=x^2$$
\begin{align}
\int\limits_{0}^{\infty} \frac{x^{a}}{1+x^{2}} \mathrm{d}x
&= \frac{1}{2} \int\limits_{0}^{\infty} z^{(a-1)/2} \frac{\mathrm{d}z}{1+z} \\
\tag{a}
&= \frac{1}{2} \mathrm{B}\left(\frac{a+1}{2}, 1-\frac{a+1}{2} \right) \\
\tag{b}
& = \frac{1}{2} \Gamma\left(\frac{a+1}{2}\right) \Gamma\left(1-\frac{a+1}{2}\right) \\
\tag{c}
&= \frac{\pi}{2\sin(\pi(a+1)/2)}
\end{align}
a. We used the following definition of the beta function

\mathrm{B}(a.b)=\int\limits_{0}^{\infty} \frac{z^{a-1}}{1+z^{a+b}} \mathrm{d}z

b. $$\mathrm{B}(a.b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
c. $$\Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin(\pi z)}$$

## Method 2

Let

f(z) = \frac{z^{a}}{1+z^{2}}

Using the keyhole contour, we have first order poles at $$\pm i$$, so the residues are

\mathrm{Res}[f(z),i] = \frac{i^{a}}{i2} = \mathrm{e}^{ia\pi/2} \frac{1}{i2}

\mathrm{Res}[f(z),-i] = \frac{(-i)^{a}}{-i2} = -\mathrm{e}^{ia\pi} \mathrm{e}^{ia\pi/2} \frac{1}{i2}

\begin{align}
\oint\limits_{C} f(z) \mathrm{d}z
&= \pi \mathrm{e}^{ia\pi/2} \left(1 – \mathrm{e}^{ia\pi} \right) \\
&= \lim_{\epsilon,R \to 0,\infty} \int\limits_{\epsilon}^{R} f(x) \mathrm{d}x
+ \int\limits_{\Gamma} f(z)\mathrm{d}z
+ \int\limits_{R}^{\epsilon} f(x) \mathrm{d}x
+ \int\limits_{\gamma} f(z)\mathrm{d}z \\
&= \int\limits_{0}^{\infty} \frac{x^{a}}{1+x^{2}} \mathrm{d}x
\,- \int\limits_{0}^{\infty} \mathrm{e}^{ia2\pi} \frac{x^{a}}{1+x^{2}} \mathrm{d}x \\
&= \left(1 – \mathrm{e}^{ia2\pi} \right) \int\limits_{0}^{\infty} \frac{x^{a}}{1+x^{2}} \mathrm{d}x
\end{align}

Thus we have
\begin{align}
\int\limits_{0}^{\infty} \frac{x^{a}}{1+x^{2}} \mathrm{d}x
&= \pi \mathrm{e}^{ia\pi/2} \left(1 – \mathrm{e}^{ia\pi} \right) \frac{1}{\left(1 – \mathrm{e}^{ia2\pi} \right)} \\
&= \frac{\pi \sin(a\pi/2)}{\sin(a\pi)} \\
&= \frac{\pi}{2\cos(a\pi/2)}
\end{align}

Notes:
1. R is the radius of the large circle $$\Gamma$$.
2. $$\epsilon$$ is the radius of the small circle $$\gamma$$.

## Integrate $$\int_{0}^{\infty} (\sqrt{x^{4}+a^{4}} + \sqrt{x^{4}+b^{4}})^{-1} \mathrm{d} x$$

How to evaluate

\int\limits_{0}^{\infty} \frac{1}{\sqrt{x^{4}+a^{4}} + \sqrt{x^{4}+b^{4}}} \mathrm{d} x
\label{eq:160824-1}
\tag{1}

was a question on Mathematics Stack Exchange. One of the answers outlined a way to convert the integral into beta functions but did not work out the solution. To do so requires a clever trick to ensure convergence of the integral, which is my reason for posting a full solution here.

We begin by multiplying the integrand by

\frac{\sqrt{x^{4}+a^{4}} – \sqrt{x^{4}+b^{4}}}{\sqrt{x^{4}+a^{4}} – \sqrt{x^{4}+b^{4}}}

so that we now have

\frac{1}{a^{4}-b^{4}} \int\limits_{0}^{\infty} \sqrt{x^{4}+a^{4}} – \sqrt{x^{4}+b^{4}} \, \mathrm{d} x
\label{eq:160824-2}
\tag{2}

The problem here is that

\int\limits_{0}^{\infty} \sqrt{x^{4}+a^{4}} \, \mathrm{d} x

does not converge. Here is where we invoke the clever trick I referred to. Let us consider

\int\limits_{0}^{\infty} (x^{4}+a^{4})^{p} \mathrm{d} x
\label{eq:160824-3}
\tag{3}

Making the successive substitutions $$x^{4} = a^{4}y^{4}$$ and $$y^{4} = z$$ yields
\begin{align}
\int\limits_{0}^{\infty} (x^{4}+a^{4})^{p} \mathrm{d} x & = \frac{a^{4p+1}}{4} \int\limits_{0}^{\infty} z^{-3/4} (z+1)^{p} \mathrm{d} z \\
& = \frac{a^{4p+1}}{4} \mathrm{B}\left(\frac{1}{4},\frac{-1}{4}-p\right)
\label{eq:160824-4}
\tag{4}
\end{align}

Putting it all together and letting $$p = 1/2$$, we have
\begin{align}
\int\limits_{0}^{\infty} \frac{1}{\sqrt{x^{4}+a^{4}} + \sqrt{x^{4}+b^{4}}} \mathrm{d} x & = \frac{1}{a^{4}-b^{4}} \int\limits_{0}^{\infty} \sqrt{x^{4}+a^{4}} – \sqrt{x^{4}+b^{4}} \mathrm{d} x \\
& = \frac{1}{4} \frac{a^{3}-b^{3}}{a^{4}-b^{4}} \mathrm{B}\left(\frac{1}{4},\frac{-3}{4}\right) \\
& \approx \frac{a^{3}-b^{3}}{a^{4}-b^{4}} 1.23605
\end{align}