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

\begin{equation}
\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
\end{equation}

\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.

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 \(\int_{0}^{1} \frac{\sqrt{1-x^{4}}}{1+x^{2}} dx\)

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

Let \(z=x^{2}\)
\begin{align}
\int\limits_{0}^{1} \frac{\sqrt{1-x^{4}}}{1+x^{2}} dx &=
\int\limits_{0}^{1} (1-x^{2})^{1/2}(1+x^{2})^{-1/2} dx \\
&= \frac{1}{2} \int\limits_{0}^{1} (1-z)^{1/2} (1+z)^{-1/2} z^{-1/2} dz \\
&= \frac{\Gamma(1/2) \Gamma(3/2)}{2 \Gamma(2)} {}_{2}\mathrm{F}_{1}(1/2,1/2;2;-1) \\
&\approx 0.7119586
\end{align}

We used Gauss’s hypergeometric function.

On The Extended Incomplete Pochhammer Symbols and Hypergeometric Functions by Rakesh Kumar Parmar, R.K. Raina

In this paper, we first introduce certain forms of extended incomplete Pochhammer symbols which are then used to define families of extended incomplete generalized hypergeometric functions. For these functions, we investigate various properties including the integral representations, derivative formula, certain generating function and fractional integrals (and derivatives) relationships. Some special cases of the main results are also deduced.

The entire paper is available here.

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 \(\int_{0}^{1} \frac{x^{n}}{\sqrt{x^{3}+1}} dx\)

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 \(\int_{0}^{\infty} \frac{x^{3}}{\mathrm{e}^{x}-1} dx\)

How to evaluate
\begin{equation}
\int\limits_{0}^{\infty} \frac{x^{3}}{\mathrm{e}^{x}-1} dx
\end{equation}
was a question posed at Mathematics Stack Exchange. Here is my solution to a more general problem.

We consider the more general integral
\begin{align}
\int\limits_{0}^{\infty} \frac{x^{n}}{\mathrm{e}^{ax}-1} dx
&= \int\limits_{0}^{\infty} \frac{\mathrm{e}^{-ax} x^{n}}{1-\mathrm{e}^{-ax}} dx \\
&= \int\limits_{0}^{\infty} \mathrm{e}^{-ax} x^{n} \sum\limits_{k = 0}^{\infty} \mathrm{e}^{-kax} dx \\
&= \sum\limits_{k = 0}^{\infty} \int\limits_{0}^{\infty} x^{n} \mathrm{e}^{-(a+ak)x} dx
\end{align}

To evaluate the integral, let \((a+ak)x = z\)
\begin{equation}
\int\limits_{0}^{\infty} x^{n} \mathrm{e}^{-(a+ak)x} dx
= \frac{1}{a^{n+1}} \frac{1}{(1+k)^{n+1}} \int\limits_{0}^{\infty} z^{n} \mathrm{e}^{-z} dz
= \frac{1}{a^{n+1}} \frac{1}{(1+k)^{n+1}} \Gamma(n+1)
\end{equation}

Now we have
\begin{equation}
\int\limits_{0}^{\infty} \frac{x^{n}}{\mathrm{e}^{ax}-1} dx
= \frac{1}{a^{n+1}} \Gamma(n+1) \sum\limits_{k = 1}^{\infty} \frac{1}{k^{n+1}}
= \frac{1}{a^{n+1}} \Gamma(n+1) \zeta(n+1)
\end{equation}

The original problem is
\begin{equation}
\int\limits_{0}^{\infty} \frac{x^{3}}{\mathrm{e}^{x}-1} dx
= \Gamma(4) \zeta(4) = 6\frac{\pi^{4}}{90} = \frac{\pi^{4}}{15}
\end{equation}

Integration Trick \([f(x)]^{-a} = \frac{1}{\Gamma(a)} \int_{0}^{\infty} z^{a-1} \mathrm{e}^{-f(x)z} \mathrm{d}z\)

A useful integration trick to convert the reciprocal of a function into an integral is
\begin{equation}
\frac{1}{[f(x)]^{a}} = \frac{1}{\Gamma(a)} \int\limits_{0}^{\infty} z^{a-1} \mathrm{e}^{-f(x)z} \mathrm{d}z
\end{equation}

The derivation is straightforward, let \(y=f(x)z\) and use the usual integral definition of the gamma function. However, one must be careful calculating the new limits of integration. If they differ from 0 and/or \(\infty\) then the incomplete gamma function must be used and the resulting integral may cause other difficulties.

See here for an excellent example of how this trick can greatly simplify the evaluation of a difficult integral.

Generalized Fresnel Integrals

The generalized fresnel integrals, also known as Böhmer’s integrals, appear in Volume 2 of Higher Transcendental Functions (Bateman Manuscript), Section 9.10, Equations 1 and 2:

\begin{align}
\tag{1a}
\label{eq:gfi-1a}
\mathrm{C}(x,a) &= \int\limits_{x}^{\infty} t^{a-1} \cos(t) \mathrm{d}t \\
\tag{1b}
\label{eq:gfi-1b}
&= \int\limits_{x}^{\infty} t^{a-1}\, \frac{\mathrm{e}^{it}+\mathrm{e}^{-it}}{2} \mathrm{d}t \\
&= \frac{1}{2} \Big[\mathrm{e}^{i\frac{\pi}{2}a} \Gamma(a,-ix) + \mathrm{e}^{-i\frac{\pi}{2}a} \Gamma(a,ix) \Big]
\tag{1c}
\label{eq:gfi-1c}
\end{align}

\begin{align}
\tag{2a}
\label{eq:gfi-2a}
\mathrm{S}(x,a) &= \int\limits_{x}^{\infty} t^{a-1} \sin(t) \mathrm{d}t \\
\tag{2b}
\label{eq:gfi-2b}
&= \int\limits_{x}^{\infty} t^{a-1}\, \frac{\mathrm{e}^{it}-\mathrm{e}^{-it}}{i2} \mathrm{d}t \\
&= \frac{1}{i2} \Big[\mathrm{e}^{i\frac{\pi}{2}a} \Gamma(a,-ix) – \mathrm{e}^{-i\frac{\pi}{2}a} \Gamma(a,ix) \Big]
\tag{2c}
\label{eq:gfi-2c}
\end{align}

To derive these results, we require a result from Volume 1 of Higher Transcendental Functions (Bateman Manuscript), Section 1.5.1:

Next consider \(\int_{C} z^{a-1} \mathrm{e}^{cz} \mathrm{d}z\) where the contour C consists of the real axis from \(+\epsilon\) to +R, the arc of the circle \(z=R\mathrm{e}^{i\phi}\) from \(\phi = 0\) to \(\phi = \beta \,\, (-\pi /2 \leq \beta \leq \pi /2)\), the straight line from \(z=R\mathrm{e}^{i\beta}\) to \(\epsilon \mathrm{e}^{i\beta}\), and the arc of the circle \(z=\epsilon \mathrm{e}^{i\phi}\) from \(\phi = \beta\) to \(\phi = 0\). Since the value of the contour integral is zero, on making \(\epsilon \to 0\) and \(R \to \infty\) it follows that
\begin{equation}
\int\limits_{0}^{\infty} t^{a-1} \mathrm{e}^{-ct\cos(\beta)\,-ict\sin(\beta)} \mathrm{d}t
= \Gamma(a) c^{-a} \mathrm{e}^{-ia\beta}
\tag{3}
\label{eq:gfi-3}
\end{equation}
for
\begin{equation}
-\frac{1}{2}\pi \lt \beta \lt \frac{1}{2}\pi ,\,\,\,\, \Re a \gt 0,\,\, \mathrm{or} \,\,\, \beta = \pm \frac{1}{2}\pi ,\,\,\,\, 0 \lt \Re a \lt 1
\end{equation}

We begin our derivation with
\begin{equation}
\int\limits_{x}^{\infty} t^{a-1}\, \mathrm{e}^{-it} \mathrm{d}t
= \int\limits_{0}^{\infty} t^{a-1}\, \mathrm{e}^{-it} \mathrm{d}t \,- \int\limits_{0}^{x} t^{a-1}\, \mathrm{e}^{-it} \mathrm{d}t
\tag{4}
\label{eq:gfi-4}
\end{equation}
for the first integral in equation \eqref{eq:gfi-4}, via the reference above, for \(\beta = \pi /2\) and \(c = 1\), equation \eqref{eq:gfi-3} becomes
\begin{equation}
\int\limits_{0}^{\infty} t^{a-1}\, \mathrm{e}^{-it} \mathrm{d}t = \Gamma(a) \mathrm{e}^{-i\frac{\pi}{2}a}
\tag{5}
\label{eq:gfi-5}
\end{equation}
for the second integral in equation \eqref{eq:gfi-4}, we use the substitution \(y=it\)
\begin{equation}
\int\limits_{0}^{x} t^{a-1}\, \mathrm{e}^{-it} \mathrm{d}t = \frac{1}{i^{a}} \int\limits_{0}^{ix} y^{a-1}\, \mathrm{e}^{-y} \mathrm{d}y = \mathrm{e}^{-i\frac{\pi}{2}a} \gamma(a,ix)
\tag{6}
\label{eq:gfi-6}
\end{equation}

Substituting equations \eqref{eq:gfi-5} and \eqref{eq:gfi-6} into equation \eqref{eq:gfi-4} yields
\begin{equation}
\int\limits_{x}^{\infty} t^{a-1}\, \mathrm{e}^{-it} \mathrm{d}t = \mathrm{e}^{-i\frac{\pi}{2}a} [\Gamma(a) \,-\, \gamma(a,ix)] = \mathrm{e}^{-i\frac{\pi}{2}a} \Gamma(a,ix)
\tag{7}
\label{eq:gfi-7}
\end{equation}

Using the same arguments, beginning with substituting \(\beta = -\pi /2\) into equation \eqref{eq:gfi-3} we have
\begin{equation}
\int\limits_{x}^{\infty} t^{a-1}\, \mathrm{e}^{it} \mathrm{d}t = \mathrm{e}^{i\frac{\pi}{2}a} [\Gamma(a) \,-\, \gamma(a,-ix)] = \mathrm{e}^{i\frac{\pi}{2}a} \Gamma(a,-ix)
\tag{8}
\label{eq:gfi-8}
\end{equation}

Substituting equations \eqref{eq:gfi-7} and \eqref{eq:gfi-8} into equations \eqref{eq:gfi-1b} and \eqref{eq:gfi-2b} yields the generalized fresnel integrals.

Notes:

  1. \(\gamma(a,x) = \int_{0}^{x} t^{a-1} \mathrm{e}^{-t} \mathrm{d}t \) is the lower incomplete gamma function.
  2. \(\Gamma(a,x) = \int_{x}^{\infty} t^{a-1} \mathrm{e}^{-t} \mathrm{d}t \) is the upper incomplete gamma function.
  3. \(\Gamma(a) = \Gamma(a,x) + \gamma(a,x)\)
  4. This analysis was necessary in order to avoid difficulties when using the upper incomplete gamma function directly with the left hand sides of equations \eqref{eq:gfi-4} and \eqref{eq:gfi-8}. Doing so results, after a change in variables, in an upper limit of integration of \(i\infty\). If one treats this as a modulus and replaces it with \(\infty\) one obtains the correct results for the generalized fresnel integrals. However, I have not been able to justify this, thus the method presented here.

Evaluate the Integral \(\int_{0}^{\infty} \frac{\ln(x)}{1+e^{ax}} \mathrm{d}x\)

How to evaluate

\begin{equation}
I = \int\limits_{0}^{\infty} \frac{\ln(x)}{1+e^{ax}} \mathrm{d}x
\tag{1}
\label{eq:161008-1}
\end{equation}
was a question posed at Mathematics Stack Exchange. As of the time of writing this post, there was a good solution posted there. Here is my solution.

Let
\begin{align}
I_{1} &= \int\limits_{0}^{\infty} \frac{x^{b}}{1+e^{ax}} \mathrm{d}x = \int\limits_{0}^{\infty} \frac{x^{b}e^{-ax}}{1+e^{-ax}} \mathrm{d}x \\
&= \sum\limits_{n=0}^{\infty} (-1)^{n} \int\limits_{0}^{\infty} x^{b} e^{-(a+na)x} \mathrm{d}x
\end{align}

We designate the last integral on the right as \(I_{2}\) and make the substitution \(y=(a+na)x\)
\begin{align}
I_{2} &= \int\limits_{0}^{\infty} x^{b} e^{-(a+na)x} \mathrm{d}x \\
&= \frac{1}{(a+na)^{b+1}} \int\limits_{0}^{\infty} y^{b} e^{-y} \mathrm{d}y \\
&= \frac{\Gamma(b+1)}{(a+na)^{b+1}} \\
&= \frac{\Gamma(b+1)}{a^{b+1}} \frac{1}{(n+1)^{b+1}}
\end{align}

Now \(I_{1}\) becomes
\begin{equation}
I_{1} = \frac{\Gamma(b+1)}{a^{b+1}} \sum\limits_{n=0}^{\infty} (-1)^{n} \frac{1}{(n+1)^{b+1}} = \frac{\Gamma(b+1)}{a^{b+1}} \eta(b+1)
\end{equation}
and we have
\begin{align}
I &= \lim_{b \to 0} \frac{\partial I_{1}}{\partial b} = \lim_{b \to 0} \int\limits_{0}^{\infty} \frac{x^{b} \ln(x)}{1+e^{ax}} \mathrm{d}x \\
&= \lim_{b \to 0} \frac{\partial}{\partial b} \frac{\Gamma(b+1)\eta(b+1)}{a^{b+1}} \\
&= \lim_{b \to 0} \frac{a^{b+1} \Big[ \Gamma(b+1)\eta^{\prime}(b+1) + \Gamma^{\prime}(b+1)\eta(b+1) \Big] – \Gamma(b+1)\eta(b+1)a^{b+1}\ln(a)}{\left(a^{b+1}\right)^{2}} \\
&= \lim_{b \to 0} \frac{\Gamma(b+1)\eta^{\prime}(b+1) + \Gamma^{\prime}(b+1)\eta(b+1) – \Gamma(b+1)\eta(b+1)\ln(a)}{a^{b+1}} \\
\tag{a}
&= \frac{1}{a} \left(\Big[\gamma \ln(2) – \frac{1}{2} \ln^{2}(2)\Big] -\gamma \ln(2) – \ln(2)\ln(a) \right) \\
&= -\frac{\ln(2)}{a} \left(\frac{1}{2} \ln(2) + \ln(a) \right)
\end{align}

In step (a) we have
\begin{align}
\lim_{b \to 0} \Gamma^{\prime}(b+1)\eta(b+1) &= \lim_{b \to 0} \Gamma^(b+1)\psi(b+1)\eta(b+1) \\
&= \Gamma(1)\psi(1)\eta(1) \\
&= -\gamma \ln(2)
\end{align}
and
\begin{align}
\lim_{b \to 0} \Gamma(b+1)\eta^{\prime}(b+1) &= \lim_{s \to 1} \eta^{\prime}(s) \\
&= \lim_{s \to 1} \sum\limits_{n=0}^{\infty} (-1)^{n} \frac{\ln(n)}{n^{s}} \\
&= \gamma \ln(2) – \frac{1}{2} \ln^{2}(2)
\end{align}
See here for a proof of this result.

Notes:

  1. \(\Gamma(z)\) is the Gamma function.
  2. \(\eta(s)\) is the Dirichlet eta function.
  3. \(\zeta(s)\) is the Riemann zeta function.
  4. \(\psi(z)\) is the digamma function.
  5. \(\gamma\) is the Euler-Mascheroni constant.