Integrals of Gradshteyn and Ryzhik: 3.267

We use the following definitions of the gamma and beta functions:
\begin{equation}
\Gamma(z) = \int\limits_{0}^{\infty} \mathrm{e}^{-t} t^{z-1} dt, \,\, \mathfrak{R}(z) \gt 0
\end{equation}

\begin{equation}
\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
\end{equation}

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

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

How to prove
\begin{equation}
\int\limits_{0}^{1} (1-x^{1/k})^{n} dx = {1\over {k+n\choose n}}
\end{equation}
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
\begin{equation}
\int \frac{x^{2-2/n}}{(3-x)^{1/n}} dx
\end{equation}
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
\begin{equation}
\int x\left(1+\frac{a}{x^{k}}\right)^{-m} \mathrm{d}x
\end{equation}
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.
\begin{equation}
\mathrm{B}_{z}(p,q) = \int_{0}^{z} t^{p-1} (1-t)^{q-1} \mathrm{d}t
\end{equation}
is the incomplete beta function.
2.
\begin{equation}
\mathrm{B}_{z}(p,q) = \frac{z^{p}}{p} \,{}_{2}\mathrm{F}_{1}(p,1-q;p+1;z)
\end{equation}
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
\begin{equation}
\int x^{km+1}(a+x^{k})^{-m} \mathrm{d}x
\end{equation}
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.
\begin{equation}
\mathrm{B}_{z}(p,q) = \int_{0}^{z} t^{p-1} (1-t)^{q-1} \mathrm{d}t
\end{equation}
is the incomplete beta function.
2.
\begin{equation}
\mathrm{B}_{z}(p,q) = \frac{z^{p}}{p} \,{}_{2}\mathrm{F}_{1}(p,1-q;p+1;z)
\end{equation}
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
\begin{equation}
\mathrm{B}(a.b)=\int\limits_{0}^{\infty} \frac{z^{a-1}}{1+z^{a+b}} \mathrm{d}z
\end{equation}
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
\begin{equation}
f(z) = \frac{z^{a}}{1+z^{2}}
\end{equation}
Using the keyhole contour, we have first order poles at \(\pm i\), so the residues are
\begin{equation}
\mathrm{Res}[f(z),i] = \frac{i^{a}}{i2} = \mathrm{e}^{ia\pi/2} \frac{1}{i2}
\end{equation}
\begin{equation}
\mathrm{Res}[f(z),-i] = \frac{(-i)^{a}}{-i2} = -\mathrm{e}^{ia\pi} \mathrm{e}^{ia\pi/2} \frac{1}{i2}
\end{equation}

\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}

Keyhole Contour. Image from https://en.wikipedia.org/wiki/Methods_of_contour_integration
Keyhole Contour. Image from https://en.wikipedia.org/wiki/Methods_of_contour_integration

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
\begin{equation}
\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}
\end{equation}

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
\begin{equation}
\frac{\sqrt{x^{4}+a^{4}} – \sqrt{x^{4}+b^{4}}}{\sqrt{x^{4}+a^{4}} – \sqrt{x^{4}+b^{4}}}
\end{equation}
so that we now have
\begin{equation}
\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}
\end{equation}

The problem here is that
\begin{equation}
\int\limits_{0}^{\infty} \sqrt{x^{4}+a^{4}} \, \mathrm{d} x
\end{equation}
does not converge. Here is where we invoke the clever trick I referred to. Let us consider
\begin{equation}
\int\limits_{0}^{\infty} (x^{4}+a^{4})^{p} \mathrm{d} x
\label{eq:160824-3}
\tag{3}
\end{equation}
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}