## Integrals of Gradshteyn and Ryzhik: 6.161 – Mellin Transforms of Theta Functions with Respect to the Lattice Parameter

We use definitions of the theta functions, shown below, from GR. Note that there is no standard notation for the theta functions.

$$z =$$ argument, $$\tau =$$ lattice parameter ($$\mathfrak{I}(\tau) \gt 0$$), and
$$q = \mathrm{e}^{i\pi \tau}$$ ($$|q| \lt 1$$)

\begin{align}
\tag{1a}
\label{eq:theta1e}
\theta_{1}(z|\tau) &= \theta_{1}(z,q) = 2 \sum_{n=0}^{\infty} (-1)^{n} q^{(n+1/2)^{2}} \sin[(2n+1)z] \\
\tag{1b}
\label{eq:theta1t}
&= -i \sum_{n=-\infty}^{\infty} (-1)^{n} q^{(n+1/2)^{2}} \mathrm{e}^{i(2n+1)z}
\end{align}

\begin{align}
\tag{2a}
\label{eq:theta2e}
\theta_{2}(z|\tau) &= \theta_{2}(z,q) = 2 \sum_{n=0}^{\infty} q^{(n+1/2)^{2}} \cos[(2n+1)z] \\
\tag{2b}
\label{eq:theta2t}
&= \sum_{n=-\infty}^{\infty} q^{(n+1/2)^{2}} \mathrm{e}^{i(2n+1)z}
\end{align}

\begin{align}
\tag{3a}
\label{eq:theta3e}
\theta_{3}(z|\tau) &= \theta_{3}(z,q) = 1 + 2 \sum_{n=1}^{\infty} q^{n^{2}} \cos(2nz) \\
\tag{3b}
\label{eq:theta3t}
&= \sum_{n=-\infty}^{\infty} q^{n^{2}} \mathrm{e}^{i2nz}
\end{align}

\begin{align}
\tag{4a}
\label{eq:theta4e}
\theta_{4}(z|\tau) &= \theta_{4}(z,q) = 1 + 2 \sum_{n=1}^{\infty} q^{n^{2}} \cos(2nz) \\
\tag{4b}
\label{eq:theta4t}
&= \sum_{n=-\infty}^{\infty} q^{n^{2}} \mathrm{e}^{i2nz}
\end{align}

## 6.161.1

\int_{0}^{\infty} x^{s-1} \theta_{2}(0|ix^2) dx
= 2\int_{0}^{\infty} x^{s-1} \sum_{n=0}^{\infty} \mathrm{e}^{-\pi x^{2}(n+1/2)^2} dx

\begin{align}
\int_{0}^{\infty} x^{s-1} \mathrm{e}^{-\pi x^{2}(n+1/2)^2} dx
&= \frac{1}{2(\pi (n+1/2)^2)^{s/2}} \int_{0}^{\infty} y^{-1+s/2} \mathrm{e}^{-y} dy \\
&= \frac{1}{2(\pi (n+1/2)^2)^{s/2}} \Gamma(s/2)
\end{align}

We used the substitution $$y = \pi x^{2}(n+1/2)^2$$.

Now we have
\begin{align}
\int_{0}^{\infty} x^{s-1} \theta_{2}(0|ix^2) dx &=
\frac{1}{\pi ^{s/2}} \Gamma(s/2) \sum_{n=0}^{\infty} \frac{1}{(n+1/2)^s} \\
&= \frac{2^s}{\pi ^{s/2}} \Gamma(s/2) \sum_{n=0}^{\infty} \frac{1}{(2n+1)^s} \\
&= \frac{2^s}{\pi ^{s/2}} \Gamma(s/2) (1-2^{-s}) \zeta(s)
\end{align}

## 6.161.2

\begin{align}
\int_{0}^{\infty} x^{s-1} [\theta_{3}(0|ix^2) \,- 1] dx
&= \int_{0}^{\infty} x^{s-1} \left(\left[1 + 2\sum_{n=1}^{\infty} \mathrm{e}^{-\pi x^{2} n^2} \right] -1 \right) dx \\
&= 2\sum_{n=1}^{\infty} \int_{0}^{\infty} x^{s-1} \mathrm{e}^{-\pi x^{2} n^2} dx \\
&= 2\sum_{n=1}^{\infty} \frac{\Gamma(s/2)}{2\pi^{s/2} n^{s}} \\
&= \frac{\Gamma(s/2)}{\pi^{s/2}} \sum_{n=1}^{\infty} \frac{1}{n^{s}} \\
&= \frac{\Gamma(s/2)}{\pi^{s/2}} \zeta(s)
\end{align}

We used the substitution $$y = \pi x^{2} n^2$$ and the same work as above to evaluate the integral.

## 6.161.3

\begin{align}
\int_{0}^{\infty} x^{s-1} [1 – \theta_{4}(0|ix^2)] dx
&= \int_{0}^{\infty} x^{s-1} \left(1 – \left[1 + 2\sum_{n=1}^{\infty} (-1)^{n} \mathrm{e}^{-\pi x^{2} n^2} \right] \right) dx \\
&= -2\sum_{n=1}^{\infty} (-1)^{n} \int_{0}^{\infty} x^{s-1} \mathrm{e}^{-\pi x^{2} n^2} dx \\
&= -2\sum_{n=1}^{\infty} (-1)^{n} \frac{\Gamma(s/2)}{2\pi^{s/2} n^{s}} \\
&= \frac{\Gamma(s/2)}{\pi^{s/2}} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{s}} \\
&= \frac{\Gamma(s/2)}{\pi^{s/2}} (1 – 2^{1-s}) \zeta(s)
\end{align}

We used the substitution $$y = \pi x^{2} n^2$$ and the same work as above to evaluate the integral.

## 6.161.4

\begin{align}
\int_{0}^{\infty} x^{s-1} [\theta_{4}(0|ix^2) + \theta_{2}(0|ix^2) \,- \theta_{3}(0|ix^2)] dx
&= \int_{0}^{\infty} x^{s-1} ([\theta_{4}(0|ix^2) \,-1] + \theta_{2}(0|ix^2) \,- [\theta_{3}(0|ix^2) \,-1]) dx \\
&= \frac{\Gamma(s/2)}{\pi^{s/2}} (2^{1-s} – 1) \zeta(s)
+ \frac{2^s}{\pi ^{s/2}} \Gamma(s/2) (1-2^{-s}) \zeta(s)
\,- \frac{\Gamma(s/2)}{\pi^{s/2}} \zeta(s) \\
&= -\frac{\Gamma(s/2)}{\pi^{s/2}} \zeta(s) [1 – 2^{1-s} – 2^{-s} + 1 + 1] \\
&= -\frac{\Gamma(s/2)}{\pi^{s/2}} \zeta(s) (2^{-s} – 1)(2^{1-s} – 1)
\end{align}
Here we used the previous 3 results.

All Riemann zeta function expressions can be found here.

## Evaluate the Integral $$\int_{0}^{\infty} \frac{\mathrm{e}^{-x}}{x^{1/2}+ax^{3/2}} \mathrm{d}x$$

How to evaluate

\int\limits_{0}^{\infty} \frac{\mathrm{e}^{-x}}{x^{1/2}+ax^{3/2}} \mathrm{d}x

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

\begin{align}
\tag{a}
\int\limits_{0}^{\infty} \frac{\mathrm{e}^{-x}}{x^{1/2}+ax^{3/2}} \mathrm{d}x &=
2\int\limits_{0}^{\infty} \frac{z\mathrm{e}^{-z^{2}}}{z+az^{3}} \mathrm{d}z \\
\tag{b}
&= \frac{2}{\sqrt{a}} \int\limits_{0}^{\infty} \mathrm{e}^{-y^{2}/a} \frac{1}{1+y^{2}} \mathrm{d}y \\
\tag{c}
&= \frac{\pi}{\sqrt{a}} \mathrm{e}^{1/a} \,\mathrm{erfc}\left(\frac{1}{\sqrt{a}}\right)
\end{align}

Notes:

a. $$x=z^{2}$$
b. $$y^{2}=az^{2}$$
c. From DLMF, we have the following integral definition of the complementary error function

\mathrm{erfc}(z) = \frac{2}{\pi} \mathrm{e}^{-z^{2}} \int\limits_{0}^{\infty} \mathrm{e}^{-z^{2}t^{2}}
\frac{1}{t^{2} + 1} \mathrm{d}t

To derive the complementary error function integral above, we begin with

\frac{2}{\pi} \mathrm{e}^{-z^{2}} \int\limits_{0}^{\infty} \frac{\mathrm{e}^{-z^{2}x^{2}}}{x^{2} + 1} \mathrm{d}x =
\frac{2z}{\pi} \mathrm{e}^{-z^{2}} \int\limits_{0}^{\infty} \frac{\mathrm{e}^{-t^{2}}}{z^{2} + t^{2}} \mathrm{d}t
= \frac{z}{\pi} \mathrm{e}^{-z^{2}} \int\limits_{-\infty}^{\infty} \frac{\mathrm{e}^{-t^{2}}}{z^{2} + t^{2}} \mathrm{d}t

We used the substitution $$x=t/z$$.

For the integral

\int\limits_{-\infty}^{\infty} \frac{\mathrm{e}^{-t^{2}}}{z^{2} + t^{2}} \mathrm{d}t

we let $$f(t) = \mathrm{e}^{-t^{2}}$$ and $$g(t) = 1/(z^{2} + t^{2})$$ and take Fourier transforms of each,

\mathrm{F}(s) = \mathcal{F}[f(t)] = \frac{\mathrm{e}^{-s^{2}/4}}{\sqrt{2}}

and

\mathrm{G}(s) = \mathcal{F}[g(t)] = \frac{1}{z}\sqrt{\frac{\pi}{2}} \mathrm{e}^{-z|s|}

then invoke Parseval’s theorem

\int\limits_{-\infty}^{\infty} f(t)\overline{g(t)} \mathrm{d}t
= \int\limits_{-\infty}^{\infty} \mathrm{F}(s)\overline{\mathrm{G}(s)} \mathrm{d}s

Dropping constants, the integral becomes
\begin{align}
\int\limits_{-\infty}^{\infty} \mathrm{e}^{-s^{2}/4} \mathrm{e}^{-z|s|} \mathrm{d}s
&= 2\int\limits_{0}^{\infty} \mathrm{e}^{-s^{2}/4} \mathrm{e}^{-z|s|} \mathrm{d}s \\
&= 2\mathrm{e}^{z^{2}} \int\limits_{0}^{\infty} \mathrm{e}^{-(s+2z)^{2}/4} \mathrm{d}s \\
&= 4\mathrm{e}^{z^{2}} \int\limits_{0}^{\infty} \mathrm{e}^{-y^{2}} \mathrm{d}y \\
&= 2\sqrt{\pi}\mathrm{e}^{z^{2}} \mathrm{erfc}(z)
\end{align}
We completed the square in the exponent and used the substitution $$y=z+s/2$$.

Putting the pieces together yields our desired result
\begin{align}
\frac{2}{\pi} \mathrm{e}^{-z^{2}} \int\limits_{0}^{\infty} \frac{\mathrm{e}^{-z^{2}x^{2}}}{x^{2} + 1} \mathrm{d}x &=
\frac{z}{\pi} \mathrm{e}^{-z^{2}} \int\limits_{-\infty}^{\infty} \frac{\mathrm{e}^{-t^{2}}}{z^{2} + t^{2}} \mathrm{d}t \\
&= \frac{z}{\pi} \mathrm{e}^{-z^{2}} \frac{1}{\sqrt{2}} \frac{1}{z} \sqrt{\frac{\pi}{2}} 2\sqrt{\pi} \mathrm{e}^{z^{2}} \mathrm{erfc}(z) \\
&= \mathrm{erfc}(z)
\end{align}

Thanks to Jack D’Aurizio for outlining this derivation.

## Evaluate the Integral $$\int_{0}^{\infty} x^{s-1} \cos(2\pi ax) \mathrm{d}x$$

How to evaluate

\int\limits_{0}^{\infty} x^{s-1} \cos(2\pi ax) \mathrm{d}x
\tag{1}
\label{eq:mtc-1}

was part of a question posed at Mathematics Stack Exchange. Note that this is the Mellin transform of the indicated cosine function.

The original answer that I provided required some rather questionable steps regarding the limits of integration, so here I provide another solution that avoids such difficulties.

In Volume 2 of Higher Transcendental Functions (Bateman Manuscript), Section 9.10, Equation 1 we have a generalization of the fresnel integrals attributed to Bohmer:
\begin{align}
\mathrm{C}(x,a) &= \int\limits_{x}^{\infty} z^{a-1} \cos(z) \mathrm{d}z \\
&= \frac{1}{2} \Big[\mathrm{e}^{i\pi a/2} \Gamma(a,-ix) + \mathrm{e}^{-i\pi a/2} \Gamma(a,ix)\Big]
\end{align}

Thus

\mathrm{C}(0,a) = \int\limits_{0}^{\infty} z^{a-1} \cos(z) \mathrm{d}z
= \Gamma(a) \cos\left(\frac{\pi}{2}a\right)

For our integral, let $$z=2\pi ax$$:

\int\limits_{0}^{\infty} x^{s-1} \cos(2\pi ax) \mathrm{d}x
= (2\pi a)^{-s} \int\limits_{0}^{\infty} z^{s-1} \cos(z) \mathrm{d}z
= (2\pi a)^{-s} \Gamma(s) \cos\left(\frac{\pi}{2}s\right)

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

How to evaluate

\int\limits_{0}^{\infty} \frac{x}{\mathrm{e}^{x}+1} \mathrm{d} x

was a question on Mathematics Stack Exchange. The original post asked for a contour integral solution and there is a good one among the answers. However, this integral can be evaluated trivially by recognizing an integral definition of the Dirichlet eta function or use of the Mellin transform.

We have

\eta(s) = \frac{1}{\Gamma(s)}\int\limits_{0}^{\infty} \frac{x^{s-1}}{\mathrm{e}^{x}+1} \mathrm{d} x

thus our integral is

\int\limits_{0}^{\infty} \frac{x}{\mathrm{e}^{x}+1} \mathrm{d} x = \eta(2)\Gamma(2) = \frac{\pi^{2}}{12}

Also
\begin{align}
\mathcal{M}\left[(\mathrm{e}^{ax}+1)^{-1}\right](s) & = \int\limits_{0}^{\infty} \frac{x^{s-1}}{\mathrm{e}^{ax}+1} \mathrm{d} x \\
& = a^{-s}\Gamma(s)(1-2^{1-s})\zeta(s) \\
& = a^{-s}\Gamma(s)\eta(s)
\end{align}
With $$a=1$$ and $$s=2$$ we arrive at our result.

## Laplace Transform of the Gudermannian Function

How to find the Laplace transform of the Gudermannian function was a question on Mathematics Stack Exchange. Here is my solution.

\mathcal{L}[\mathrm{gd}(x)](s) = \int\limits_{0}^{\infty} \mathrm{e}^{-sx}\mathrm{gd}(x)\mathrm{d} x

Integration by parts yields
\begin{align}
\int\limits_{0}^{\infty} \mathrm{e}^{-sx}\mathrm{gd}(x)\mathrm{d} x & =
\frac{-1}{s}\mathrm{gd}(x)\mathrm{e}^{-sx}\big|_{0}^{\infty} +
\frac{1}{s}\int\limits_{0}^{\infty} \mathrm{e}^{-sx}\mathrm{sech}(x)\mathrm{d} x \\
& = 0 + \frac{1}{s}\mathcal{L}[\mathrm{sech}(x)](s) \\
& = \frac{1}{2s}\left[\psi\left(\frac{s+3}{4}\right) – \psi\left(\frac{s+1}{4}\right) \right]
\end{align}
for $$Re(s) > 0$$ due to evaluating the limit at $$x = \infty$$, while $$Re(s) > -1$$ for the Laplace transform of the hyperbolic secant.

Notes:

1. $$\frac{\mathrm{d}}{\mathrm{d}x}\mathrm{gd}(x) = \mathrm{sech}(x)$$
2. $$\mathrm{gd}(0) = 0\,$$ and $$\,\mathrm{gd}(\infty) = \frac{\pi}{2}$$
3. $$\psi(s)$$ is the digamma function.

## Integrals from Blagouchine’s Malmsten Integral Paper

Here I evaluate problem 18a from page 43 from Rediscovery of Malmsten’s integrals, their evaluation by contour integration methods and some related results by Iaroslav V. Blagouchine as well as a bonus integral. The integral in question is

\int\limits_{0}^{\infty} \frac{x^{a}\mathrm{ln}(x)}{\mathrm{e}^{bx}-1} \mathrm{d} x
\label{eq:160817-1}
\tag{1}

We will evaluate this integral using two methods. First, taking the hint given by Blagouchine, we expand the function

f(x) = \frac{1}{\mathrm{e}^{bx}-1}
\label{eq:160817-2}
\tag{2}

and then integrate term by term.

f(x) = \frac{1}{\mathrm{e}^{bx}-1} = \frac{\mathrm{e}^{-bx}}{1-\mathrm{e}^{-bx}} = \sum\limits_{n=0}^{\infty} \mathrm{e}^{-(1+n)bx}
\label{eq:160817-3}
\tag{3}

We commence with

I = \int\limits_{0}^{\infty} \frac{x^{a}}{\mathrm{e}^{bx}-1} \mathrm{d} x = \int\limits_{0}^{\infty} x^{a} \sum\limits_{n=0}^{\infty} \mathrm{e}^{-(1+n)bx} \mathrm{d} x
= \sum\limits_{n=0}^{\infty} \int\limits_{0}^{\infty} x^{a} \mathrm{e}^{-(1+n)bx} \mathrm{d} x
\label{eq:160817-4}
\tag{4}

Making the substitution $$y = (1+n)bx$$ yields
\begin{align}
I & = \sum\limits_{n=0}^{\infty} \frac{1}{(1+n)^{a+1}b^{a+1}} \int\limits_{0}^{\infty} y^{a} \mathrm{e}^{-y} \mathrm{d} y
= \frac{\Gamma(a+1)}{b^{\,a+1}} \sum\limits_{n=0}^{\infty} \frac{1}{(1+n)^{a+1}} \\
& = \frac{\Gamma(a+1)}{b^{\,a+1}} \sum\limits_{k=1}^{\infty} \frac{1}{k^{a+1}} \\
& = \frac{\Gamma(a+1)\zeta(a+1)}{b^{\,a+1}}
\label{eq:160817-5}
\tag{5}
\end{align}

So we have our first result

\int\limits_{0}^{\infty} \frac{x^{a}}{\mathrm{e}^{bx}-1} \mathrm{d} x = \frac{\Gamma(a+1)\zeta(a+1)}{b^{\,a+1}}
\label{eq:160817-6}
\tag{6}

We can evaluate this integral quite simply if we recognize that it can be written as a Mellin transform

\mathcal{M}[f(x)](s) = \int\limits_{0}^{\infty} \frac{x^{s-1}}{\mathrm{e}^{bx}-1} \mathrm{d} x = \frac{\Gamma(s)\zeta(s)}{b^{s}}
\label{eq:160817-7}
\tag{7}

Letting $$s = a+1$$ yields our result.

To evaluate the integral in \eqref{eq:160817-1} we differentiate equation \eqref{eq:160817-6} with respect to $$a$$ and note that

\frac{d\Gamma(z)}{dz} = \Gamma(z)\psi^{(0)}(z)

\begin{align}
\int\limits_{0}^{\infty} \frac{x^{a}\mathrm{ln}(x)}{\mathrm{e}^{bx}-1} \mathrm{d} x & = \frac{-\mathrm{ln}(b)}{b^{\,a+1}} \Gamma(a+1)\zeta(a+1) + \frac{\Gamma'(a+1)\zeta(a+1)}{b^{\,a+1}} + \frac{\Gamma(a+1)\zeta'(a+1)}{b^{\,a+1}} \\
& = \frac{\Gamma(a+1)}{b^{\,a+1}} \left[ \psi^{(0)}(a+1)\zeta(a+1) + \zeta'(a+1) – \zeta(a+1)\mathrm{ln}(b) \right]
\label{eq:160817-8}
\tag{8}
\end{align}

Note that $$\mathrm{Re}(a) > 0$$.

## Integrate $$\int_{0}^{\infty} \frac{\mathrm{ln}^{3}(x)}{(1+x^{2})(1+x)^{2}} \mathrm{d} x$$

The following integral was a question on Mathematics Stack Exchange.

\int\limits_{0}^{\infty} \frac{\mathrm{ln}^{3}(x)}{(1+x^{2})(1+x)^{2}} \mathrm{d} x
\label{eq:160813a1}
\tag{1}

As usual, there are multiple clever solutions. However, a user noted that the integral could be evaluated via the Mellin transform but he did not provide any details so I will do it here.

Let us evaluate it via the Mellin transform

\mathcal{M}[f(x)](s) = \int\limits_{0}^{\infty} x^{s-1} f(x) \mathrm{d} x
\label{eq:160813a2}
\tag{2}

where

f(x) = \frac{1}{(1+x^{2})(1+x)^{2}} = -\frac{1}{2}\frac{x}{x^{2}+1} + \frac{1}{2}\frac{1}{x+1} + \frac{1}{2}\frac{1}{(x+1)^{2}}
\label{eq:160813a3}
\tag{3}

via partial fraction expansion.

Applying the Mellin transform, yields
\begin{align}
\mathcal{M}[f(x)](s) & = \int\limits_{0}^{\infty} \frac{x^{s-1}}{(1+x^{2})(1+x)^{2}} \\
& = -\frac{1}{2}\left[\frac{1}{2}\pi\sec\left(\frac{\pi}{2}s\right)\right] + \frac{1}{2}\pi\csc(\pi s) + \frac{1}{2} \mathrm{B}(s,2-s)
\label{eq:160813a4}
\tag{4}
\end{align}

Taking the 3rd derivative of equation \eqref{eq:160813a4} with respect to s and then taking $$\lim s \to 1$$ yields
\begin{align}
\int\limits_{0}^{\infty} \frac{\mathrm{ln}^{3}(x)}{(1+x^{2})(1+x)^{2}} \mathrm{d} x & = -\frac{1}{2}\left( -\frac{7}{960} \pi^{4} \right) + \frac{1}{2} \left( -\frac{7}{60} \pi^{4} \right) + \frac{1}{2}0 \\
& = -\frac{7}{128} \pi^{4}
\label{eq:160813a5}
\tag{5}
\end{align}

Let us fill in the details. Handling the beta function first, we have

\mathrm{B}(s,2-s) = \frac{\Gamma(s)\Gamma(2-s)}{\Gamma(2)} = \Gamma(s)\Gamma(2-s)
\label{eq:160813a6}
\tag{6}

To take derivatives, we note that

\frac{\mathrm d}{\mathrm d s} \Gamma(s) = \Gamma(s) \psi^{(0)}(s) \quad \mathrm{and} \quad \psi^{(n)}(s) = \frac{\mathrm{d}^{n}}{\mathrm{d} s^{n}} \psi^{(0)}(s)

Where $$\psi^{(n)}(s)$$ is the polygamma function.

Taking the third derivative of equation \eqref{eq:160813a6} and letting $$\lim s \to 1$$ equals 0. Here we used

\psi^{(0)}(1) = -\gamma \quad \mathrm{and} \quad \psi^{(1)}(1) = \frac{\pi^{2}}{6}

and fortunately $$\Gamma^{(3)}(s) = -\Gamma^{(3)}(2-s)$$ which leads to some cancellations.

Doing the same for the first two terms on the right hand side of equation \eqref{eq:160813a4}, we have to be careful. Each of them individually goes to $$\infty$$ as $$\lim s \to 1$$ but the $$(s-1)^{-4}$$ terms in the Laurent expansions about $$s=1$$ cancel. Here I used Wolfram Alpha. Doing it with the two terms combined yielded our final answer.

## Integrate $$\int_{0}^{\infty} \mathrm{ln}^{2}(x)\,(1+x^{2})^{-1} \mathrm{d} x$$

The following integral was a question on Mathematics Stack Exchange

\int\limits_{0}^{\infty} \frac{\mathrm{ln}^{2}(x)}{1+x^{2}} \mathrm{d} x
\label{eq:1608131}
\tag{1}

Let us evaluate it via the Mellin transform

\mathcal{M}[f(x)](s) = \int\limits_{0}^{\infty} x^{s-1} f(x) \mathrm{d} x
\label{eq:1608132}
\tag{2}

where

f(x) = \frac{1}{1+x^{2}}
\label{eq:1608133}
\tag{3}

Applying the Mellin transform, yields

\mathcal{M}[f(x)](s) = \int\limits_{0}^{\infty} \frac{x^{s-1}}{1+x^{2}} \mathrm{d} x = \frac{\pi}{2}\csc\left(\frac{\pi}{2}s\right)
\label{eq:1608134}
\tag{4}

And thus
\begin{align}
\int\limits_{0}^{\infty} \frac{\mathrm{ln}^{2}(x)}{1+x^{2}} \mathrm{d} x & = \frac{\mathrm{d}^{2}}{\mathrm{d}s^{2}} \int\limits_{0}^{\infty} \frac{x^{s-1}}{1+x^{2}} \mathrm{d} x |_{s=1} \\
& = \frac{\mathrm{d}^{2}}{\mathrm{d}s^{2}} \frac{\pi}{2}\csc\left(\frac{\pi}{2}s\right) |_{s=1} \\
& = \frac{\pi^{3}}{16}[\cos(\pi s) + 3]\csc^{3}\left(\frac{\pi}{2}s\right) |_{s=1} \\
& = \frac{\pi^{3}}{8}
\label{eq:1608135}
\tag{5}
\end{align}

## Integrate $$\int_{0}^{1} \frac{\mathrm{ln}^{3}(x)}{1+x} \mathrm{d} x$$

The following integral was a question on Mathematics Stack Exchange

\int\limits_{0}^{1} \frac{\mathrm{ln}^{3}(x)}{1+x} \mathrm{d} x
\label{eq:160812a1}
\tag{1}

We begin with the substitution $$x=\mathrm{e}^{-y}$$ so our integral becomes

-\int\limits_{0}^{\infty} \frac{y^{3}}{1+\mathrm{e}^{\,y}} \mathrm{d} y
\label{eq:160812a2}
\tag{2}

This integral is a Mellin transform of the function

f(y) = \frac{1}{1+\mathrm{e}^{\,y}}
\label{eq:160812a3}
\tag{3}

where the Mellin transform is defined as

\mathcal{M}[f(x)](s) = \int\limits_{0}^{\infty} x^{s-1} f(x) \mathrm{d} x
\label{eq:160812a4}
\tag{4}

From Tables of Integral Transforms (Bateman Manuscript) Volume 1, we have

\mathcal{M}[(\mathrm{e}^{\alpha x} + 1)^{-1}](s) = \frac{1}{\alpha^{s}} \Gamma(s) (1-2^{1-s}) \zeta(s)
\label{eq:160812a5}
\tag{5}

With $$s = 4$$ and $$\alpha = 1$$ we have

-\int\limits_{0}^{\infty} \frac{y^{3}}{1+\mathrm{e}^{\,y}} \mathrm{d} y = -\Gamma(4) (1-2^{1-4}) \zeta(4) = -\frac{7\pi^{4}}{120}
\label{eq:160812a6}
\tag{6}

Thus

\int\limits_{0}^{1} \frac{\mathrm{ln}^{3}(x)}{1+x} \mathrm{d} x = -\frac{7\pi^{4}}{120}
\label{eq:160812a7}
\tag{7}

## Addendum

We can rewrite \eqref{eq:160812a5} as
\begin{align}
\mathcal{M}[(\mathrm{e}^{\alpha x} + 1)^{-1}](s) & = \frac{1}{\alpha^{s}} \Gamma(s) (1-2^{1-s}) \zeta(s) \\
& = \frac{1}{\alpha^{s}} \Gamma(s) \eta(s)
\label{eq:160812a8}
\tag{8}
\end{align}
where

\eta(s) = (1-2^{1-s}) \zeta(s) = \sum\limits_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{s}}

is the Dirichlet eta function, also known as the alternating Riemann zeta function.

Note that for Wolfram Alpha, $$\eta(s)$$ defaults to the Dedekind eta function. To obtain the Dirichlet eta function type “dirichlet eta(s)”.

## A Laplace Transform Proof of $$\mathrm{B}(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$

This proof appeared in Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals by George Boros and Victor Moll. Here, I fill in the steps of the proof.

We begin with 3 fundamental results of the Laplace transform: the basic definition, convolution, and the Laplace transform of a convolution.

\mathcal{L}[f(t)](s) = \int\limits_{0}^{\infty} \mathrm{e}^{-st} f(t) \mathrm{d} t
\label{eq:1608061}
\tag{1}

(f*g)(t) = \int\limits_{0}^{t} f(\tau) g(t-\tau) \mathrm{d} \tau
\label{eq:1608062}
\tag{2}

\mathcal{L}[f*g] = \mathcal{L}[f]\mathcal{L}[g]
\label{eq:1608063}
\tag{3}

We begin with the following two functions

f(t) = t^{x-1} \quad \mathrm{and} \quad g(t) = t^{y-1}
\label{eq:1608064}
\tag{4}

These functions were chosen for two reasons. First, they appear in the basic integral definition of the beta function

\mathrm{B}(x,y) = \int\limits_{0}^{1} t^{x-1} (1-t)^{y-1} \mathrm{d} t
\label{eq:1608065}
\tag{5}

and, as we now show, their Laplace transforms result in gamma functions
\begin{align}
\mathcal{L}[f(t)] & = \mathcal{L}[t^{x-1}] = \int\limits_{0}^{\infty} \mathrm{e}^{-st} t^{x-1} \mathrm{d} t \\
& = \frac{1}{s^{x}} \int\limits_{0}^{\infty} \mathrm{e}^{-v} v^{x-1} \mathrm{d} v \\
& = \frac{\Gamma(x)}{s^{x}}
\label{eq:1608066}
\tag{6}
\end{align}
where we have used the substitution $$v = st$$. Likewise, taking the Laplace transform of $$g(t)$$ yields,

\mathcal{L}[g(t)] = \mathcal{L}[t^{y-1}] = \frac{\Gamma(y)}{s^{y}}
\label{eq:1608067}
\tag{7}

Now we substitute equations \eqref{eq:1608067}, \eqref{eq:1608066}, and \eqref{eq:1608062} into \eqref{eq:1608063}

\frac{\Gamma(x)}{s^{x}} \frac{\Gamma(y)}{s^{y}} = \mathcal{L}\left[\int\limits_{0}^{t} \tau^{x-1} (t – \tau)^{y-1} \mathrm{d} \tau \right]
\label{eq:1608068}
\tag{8}

To evaluate the integral, we let $$\tau = tu$$
\begin{align}
\int\limits_{0}^{t} \tau^{x-1} (t – \tau)^{y-1} \mathrm{d} \tau & = t^{x+y-1} \int\limits_{0}^{1} u^{x-1} (1 – u)^{y-1} \mathrm{d} u \\
& = t^{x+y-1} \mathrm{B}(x,y)
\label{eq:1608069}
\tag{9}
\end{align}

Thus, we have
\begin{align}
\mathcal{L}[t^{x+y-1} \mathrm{B}(x,y)] & = \frac{\Gamma(x)}{s^{x}} \frac{\Gamma(y)}{s^{y}} \\
& = \mathrm{B}(x,y) \int\limits_{0}^{\infty} \mathrm{e}^{-st} t^{x+y-1} \mathrm{d} t \\
& = \mathrm{B}(x,y) \frac{1}{s^{x+y}} \int\limits_{0}^{\infty} \mathrm{e}^{-v} v^{x+y-1} \mathrm{d} v \\
& = \mathrm{B}(x,y) \frac{\Gamma(x+y)}{s^{x+y}}
\label{eq:16080610}
\tag{10}
\end{align}

Combining the first and last results from the right hand side of the above equation yields our final result

\mathrm{B}(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}