Evaluate the Integral \(\int \frac{x}{\mathrm{e}^{x}-1} dx\)

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

\begin{equation}
I = \int \frac{x}{\mathrm{e}^{x}-1} dx = -I_{1} = -\int \frac{x}{1-\mathrm{e}^{x}} dx
\end{equation}

Integrate by parts
\begin{align}
I_{1} &= \int \frac{x}{1-\mathrm{e}^{x}} dx \\
\tag{1}
&= x^{2} – x \ln(1-\mathrm{e}^{x}) – \int x dx + \int \ln(1-\mathrm{e}^{x}) dx \\
\tag{2}
&= x^{2} – x \ln(1-\mathrm{e}^{x}) – \frac{x^{2}}{2} – \mathrm{Li}_{2}(\mathrm{e}^{x})
\end{align}
and thus
\begin{equation}
\int \frac{x}{\mathrm{e}^{x}-1} dx = x \ln(1-\mathrm{e}^{x}) – \frac{x^{2}}{2} + \mathrm{Li}_{2}(\mathrm{e}^{x})
\end{equation}

1. Let \(u=\mathrm{e}^{x}\)
\begin{align}
\int \frac{1}{1-\mathrm{e}^{x}} dx &= \int \frac{1}{u(1-u)} du \\
&= \int \left( \frac{1}{u} + \frac{1}{1-u} \right) du \\
&= \ln \frac{u}{1-u} \\
&= x – \ln(1-\mathrm{e}^{x})
\end{align}

2. Let \(u=\mathrm{e}^{x}\)
\begin{align}
\int \ln(1-\mathrm{e}^{x}) dx &= \int \frac{\ln(1-u)}{u} du \\
&= -\mathrm{Li}_{2}(u) \\
&= -\mathrm{Li}_{2}(\mathrm{e}^{x})
\end{align}
where
\begin{equation}
\mathrm{Li}_{2}(z) = -\int\limits_{0}^{z} \frac{\ln(1-x)}{x} dx
\end{equation}
is the dilogarithm 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 the Integral \(\int_{1}^{x} \frac{dt}{\sqrt{t^{3}-1}} \)

How to evalute
\begin{equation}
\int\limits_{1}^{x} \frac{dt}{\sqrt{t^{3}-1}}
\end{equation}
was a question posed at Mathematics Stack Exchange. Here is my solution

The integral is of the form of the polynomial under the radical having 1 real and 2 complex roots and the lower limit of integration equal to the real root. So we have
\begin{equation}
\int\limits_{a}^{x} \left( (t-a)[(t-r)^{2} + s^{2}] \right)^{-1/2}\, dt
= \frac{1}{\sqrt{m}} \mathrm{cn}^{-1}\left(\frac{m-(t-a)}{m+(t-a)},k \right)
\end{equation}
where
\begin{equation}
m^{2} = (r – a)^{2} + s^{2} \quad \mathrm{and} \quad k^{2} = \frac{1}{2} – \frac{a-r}{2m}
\end{equation}
\(\mathrm{cn}^{-1}z\) is the inverse of one of the Jacobi elliptic functions and \(k\) is the modulus.

To factor the polynomial inside of the radical, we note that the roots are the cube roots of 1, thus
\begin{align}
t^{3}-1 &= (t-t_{1})(t-t_{2})(t-t_{3}) \\
&= (t-1) \left(\Big[t+\frac{1}{2}\Big]-i\frac{\sqrt{3}}{2} \right) \left(\Big[t+\frac{1}{2}\Big]+i\frac{\sqrt{3}}{2} \right) \\
&= (t-1) \left(\Big[t+\frac{1}{2}\Big]^{2} + \frac{3}{4}\right)
\end{align}

We now have
\begin{equation}
a = 1 \quad r = -\frac{1}{2} \quad s = \frac{\sqrt{3}}{2} \quad m^{2} = 3 \quad k^{2} = \frac{1}{2} – \frac{3}{4\sqrt{3}}
\end{equation}
and thus
\begin{equation}
\int\limits_{1}^{x} \frac{dt}{\sqrt{t^{3}-1}}
= 3^{-1/4} \mathrm{cn}^{-1}\left(\frac{\sqrt{3} + 1 – x}{\sqrt{3} – 1 + x},\sqrt{\frac{1}{2} – \frac{3}{4\sqrt{3}}} \right)
\end{equation}

Evaluate the Integral \( \int x^{2} \mathrm{e}^{-x^{2}} dx \)

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

Integrate by parts

\begin{equation}
\int x^{2} \mathrm{e}^{-x^{2}} dx = \frac{1}{2} \sqrt{\pi} x^{2} \mathrm{erf}(x) – \sqrt{\pi} \int x \, \mathrm{erf}(x) dx
\end{equation}

Integrate by parts again
\begin{align}
\int x \,\mathrm{erf}(x) dx
&= x^{2} \mathrm{erf}(x) + \frac{1}{\sqrt{\pi}} x \mathrm{e}^{-x^{2}}
– \int x \,\mathrm{erf}(x) dx – \frac{1}{\sqrt{\pi}} \int \mathrm{e}^{-x^{2}} dx \\
&= x^{2} \mathrm{erf}(x) + \frac{1}{\sqrt{\pi}} x \,\mathrm{e}^{-x^{2}}
– \int x \mathrm{erf}(x) dx – \frac{1}{2} \mathrm{erf}(x) \\
&= \frac{1}{2} x^{2} \mathrm{erf}(x) + \frac{1}{2\sqrt{\pi}} x \mathrm{e}^{-x^{2}}
– \frac{1}{4} \mathrm{erf}(x)
\end{align}

Thus we have
\begin{equation}
\int x^{2} \mathrm{e}^{-x^{2}} dx = \frac{\sqrt{\pi}}{4} \mathrm{erf}(x) – \frac{1}{2} x \mathrm{e}^{-x^{2}}
\end{equation}

Evaluate the Integral \( \int_{0}^{1} \mathrm{e}^{-x^{4}} (1-x^{4}) dx \)

How to evaluate
\begin{equation}
\int\limits_{0}^{1} \mathrm{e}^{-x^{4}} (1-x^{4}) dx
\end{equation}
was a question posed at Mathematics Stack Exchange. Here is my solution.

Let \(y=x^{4}\)
\begin{equation}
\int\limits_{0}^{1} \mathrm{e}^{-x^{4}} dx = \frac{1}{4} \int\limits_{0}^{1} \mathrm{e}^{-y} y^{-3/4} dy
= \frac{1}{4} \gamma\left(\frac{1}{4},1 \right)
\end{equation}

Using the same substitution, we also have
\begin{equation}
\int\limits_{0}^{1} \mathrm{e}^{-x^{4}} x^{4} dx = \frac{1}{4} \int\limits_{0}^{1} \mathrm{e}^{-y} y^{1/4} dy
= \frac{1}{4} \gamma\left(\frac{5}{4},1 \right)
\end{equation}

Thus we obtain
\begin{equation}
\int\limits_{0}^{1} \mathrm{e}^{-x^{4}} (1-x^{4}) dx
= \frac{1}{4} \Big[ \gamma\left(\frac{1}{4},1 \right) – \gamma\left(\frac{5}{4},1 \right) \Big]
\approx 0.7256
\end{equation}

We have used the lower incomplete gamma function:
\begin{equation}
\gamma(s,z) = \int\limits_{0}^{z} \mathrm{e}^{-x} x^{s-1} dx
\end{equation}

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}