## Prove $$\int_{-\infty}^{\infty} \mathrm{e}^{ist^{2}} dt = \mathrm{e}^{is\pi /4}\sqrt{\pi}$$ for $$s = \pm 1$$

First integral, for $$s=1$$
\begin{align}
2\int \mathrm{e}^{it^{2}} dt &= 2\frac{1}{\sqrt{i}} \int \mathrm{e}^{x^{2}} dx \\
&= \frac{1}{\sqrt{i}} \sqrt{\pi} \mathrm{erfi}(x) \\
&= \frac{1}{\sqrt{i}} \sqrt{\pi} \mathrm{erfi}(t\sqrt{i})
\end{align}

Thus
\begin{align}
2\int\limits_{0}^{\infty} \mathrm{e}^{it^{2}} dt &= \frac{1}{\sqrt{i}} \sqrt{\pi} \mathrm{erfi}(t\sqrt{i}) \Big|_{0}^{\infty} \\
&= \frac{1}{\sqrt{i}} \sqrt{\pi} (i-0) = \sqrt{i}\sqrt{\pi} = \mathrm{e}^{i\pi /4}\sqrt{\pi}
\end{align}

Second integral, for $$s=-1$$
\begin{align}
2\int \mathrm{e}^{-it^{2}} dt &= 2\frac{1}{\sqrt{i}} \int \mathrm{e}^{-x^{2}} dx \\
&= \frac{1}{\sqrt{i}} \sqrt{\pi} \mathrm{erf}(x) \\
&= \frac{1}{\sqrt{i}} \sqrt{\pi} \mathrm{erf}(t\sqrt{i})
\end{align}

Thus
\begin{align}
2\int\limits_{0}^{\infty} \mathrm{e}^{-it^{2}} dt &= \frac{1}{\sqrt{i}} \sqrt{\pi} \mathrm{erf}(t\sqrt{i}) \Big|_{0}^{\infty} \\
&= \frac{1}{\sqrt{i}} \sqrt{\pi} (1-0) = \frac{1}{\sqrt{i}} \sqrt{\pi} = \mathrm{e}^{-i\pi /4}\sqrt{\pi}
\end{align}

And we have
\begin{equation}
\int\limits_{-\infty}^{\infty} \mathrm{e}^{ist^{2}} dt = \mathrm{e}^{is\pi /4}\sqrt{\pi}
\end{equation}
for $$s = \pm 1$$

## Evaluate the Integral $$\int_{0}^{1} \frac{1}{1-\log_{2}x} dx$$

How to evaluate
\begin{equation}
\int\limits_{0}^{1} \frac{1}{1-\log_{2}x} dx
\end{equation}
was a question posed at Mathematics Stack Exchange. Here is my solution.

Make successive substitutions $$z=1-\frac{\ln x}{\ln2}, \,y=z\ln2$$

\begin{align}
\int\limits_{0}^{1} \frac{1}{1-\log_{2}x} dx &= 2\ln2 \int\limits_{1}^{\infty} \frac{1}{z} \mathrm{e}^{-z\ln2} dz \\
&= 2\ln2 \int\limits_{\ln2}^{\infty} \frac{\mathrm{e}^{-y}}{y} dy \\
&=-(2\ln2)\mathrm{Ei}(-\ln2) \approx 0.52495
\end{align}

\begin{equation}
\mathrm{Ei}(z) = -\int\limits_{-z}^{\infty} \frac{\mathrm{e}^{-t}}{t} dt
\end{equation}
is the exponential integral function.

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

How to prove
\begin{equation}
\lim_{n \to \infty} \int\limits_{0}^{1} \frac{x^{n}}{\sqrt{1+x^{n}}} dx = 0
\end{equation}
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
\begin{equation}
{}_{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
\end{equation}
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
\begin{equation}
\lim_{n \to \infty} \int\limits_{0}^{1} \frac{x^{n}}{\sqrt{1+x^{n}}} dx = 0*2(\sqrt{2} – 1) = 0
\end{equation}

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

How to evaluate
\begin{equation}
\int\limits_{0}^{1} \frac{x^{n}}{\sqrt{x^{3}+1}} dx
\end{equation}
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
\begin{equation}
{}_{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
\end{equation}
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
\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.

## Given $$f(x) = \int_{1}^{\sqrt{x}} \mathrm{e}^{-t^{2}} dt$$ Find $$\int_{0}^{1} \frac{f(x)}{\sqrt{x}} dx$$

This question appeared on Mathematics Stack Exchange. My solution is an alternative to the better solution here.

\begin{equation}
f(x) = \int\limits_{1}^{\sqrt{x}} \mathrm{e}^{-t^{2}} dt
\label{eq:161108-1}
\tag{1}
\end{equation}
\begin{equation}
\int\limits_{0}^{1} \frac{f(x)}{\sqrt{x}} dx
\label{eq:161108-2}
\tag{2}
\end{equation}

We need the following result
\begin{equation}
\int \mathrm{erf}(x) dx = x\,\mathrm{erf}(x) + \frac{\mathrm{e}^{-x^{2}}}{\sqrt{\pi}}
\label{eq:161108-3}
\tag{3}
\end{equation}
Proof: Integrate by parts
\begin{align}
\int \mathrm{erf}(x) dx &= x\,\mathrm{erf}(x) -\frac{2}{\sqrt{\pi}} \int x\,\mathrm{e}^{-x^{2}} dx \\
&= x\,\mathrm{erf}(x) + \frac{\mathrm{e}^{-x^{2}}}{\sqrt{\pi}}
\end{align}
we used the substitution $$u=x^{2}$$.

Now we evaluate equation \eqref{eq:161108-1}
\begin{equation}
f(x) = \frac{\sqrt{\pi}}{2} \mathrm{erf}(x) \Big|_{1}^{\sqrt{x}} = \frac{\sqrt{\pi}}{2} [\mathrm{erf}(\sqrt{x}) – \mathrm{erf}(1)]
\label{eq:161108-4}
\tag{4}
\end{equation}

Substitute equation \eqref{eq:161108-4} into equation \eqref{eq:161108-2} and evaluate the following two integrals.

\begin{align}
I_{1} &= \int\limits_{0}^{1} \frac{\mathrm{erf}(\sqrt{x})}{\sqrt{x}} dx \\
&= 2\int\limits_{0}^{1} \mathrm{erf}(z) dz \\
&= 2\,\mathrm{erf}(1) + \frac{2}{\mathrm{e}\sqrt{\pi}} – \frac{2}{\sqrt{\pi}}
\end{align}
we used the substitution $$z=\sqrt{x}$$.

\begin{equation}
I_{2} = \int\limits_{0}^{1} \frac{1}{\sqrt{x}} dx = 2
\end{equation}

Putting all of the pieces together yields our final result
\begin{align}
\int\limits_{0}^{1} \frac{f(x)}{\sqrt{x}} dx
&= \frac{\sqrt{\pi}}{2} \left(2\,\mathrm{erf}(1) + \frac{2}{\mathrm{e}\sqrt{\pi}} – \frac{2}{\sqrt{\pi}} – 2\,\mathrm{erf}(1) \right) \\
&= \frac{1}{\mathrm{e}} – 1
\end{align}

## A Proof of the Orthogonality of the Legendre Polynomials

This post was prompted by this question at Mathematics Stack Exchange. This proof of the orthogonality of the Legendre polynomials is from Special Functions and Their Applications by N. N. Lebedev, a book that I highly recommend.

We begin with Legendre’s differential equation
\begin{equation}
[(1-x^{2})P^{\prime}_{n}(x)]^{\prime} +n(n+1)P_{n}(x) = 0,\quad n \in \mathbb{Z}_{0}^{+}
\label{eq:lp1}
\tag{1}
\end{equation}

The first step is to multiple equation \eqref{eq:lp1} by $$P_{m}(x)$$ and subtract it from equation \eqref{eq:lp1} written for $$m$$ and multiplied by $$P_{n}(x)$$.

\begin{equation}
[(1-x^{2})P^{\prime}_{m}(x)]^{\prime}P_{n}(x) \,-\, [(1-x^{2})P^{\prime}_{n}(x)]^{\prime}P_{m}(x) + [m(m+1)-n(n+1)]P_{m}(x)P_{n}(x) = 0
\end{equation}

Rearrangement yields
\begin{equation}
\{(1-x^{2})[P^{\prime}_{m}(x)P_{n}(x)-P^{\prime}_{n}(x)P_{m}(x)]\}^{\prime} + (m-n)(m+n+1)P_{m}(x)P_{n}(x) = 0
\label{eq:lp2}
\tag{2}
\end{equation}

Integrating equation \eqref{eq:lp2} from -1 to 1, the first term goes to 0 and we are left with
\begin{equation}
(m-n)(m+n+1) \int\limits_{-1}^{1} P_{m}(x)P_{n}(x) dx = 0
\end{equation}
or
\begin{equation}
\int\limits_{-1}^{1} P_{m}(x)P_{n}(x) dx = 0, \quad m \ne n
\end{equation}

## Integration by differentiation: new proofs, methods and examples by Ding Jia, Eugene Tang, Achim Kempf

Recently, new methods were introduced which allow one to solve ordinary integrals by performing only derivatives. These studies were originally motivated by the difficulties of the quantum field theoretic path integral, and correspondingly, the results were derived by heuristic methods. Here, we give rigorous proofs for the methods to hold on fully specified function spaces. We then illustrate the efficacy of the new methods by applying them to the study of the surprising behavior of so-called Borwein integrals.

The entire paper is available here.