The Laplace transform of $\frac{\ln(1+at)}{1+t}$

By expressing the square of the exponential integral as a double integral and then making a change of variables, one can show $$ \int_{0}^{\infty} e^{-2zt} \ \frac{\ln(1+2t)}{1+t} \, dt = \frac{e^{2z} [\text{Ei}(-z)]^{2}}{2} \ , \ |\text{arg}(z)| \le \frac{\pi}{2}. $$ I will show this at the end of my post.

But can $$ \int_{0}^{\infty} e^{-2zt} \ \frac{\ln(1+at)}{1+t} \, dt$$ be evaluated in terms of the exponential integral or other nonelementary functions for other positive values of $a$ besides $a=2$? In particular, what about $a=1$?

$$ \begin{align} [\text{Ei}(-z)]^{2} &= \left( e^{-z} \int_{0}^{\infty} \frac{e^{-zx}}{1+x} \, dx \right) \left( \ e^{-z} \int_{0}^{\infty} \frac{e^{-zt}}{1+t} \, dt \right) \\ &= e^{-2z} \int_{0}^{\infty} \int_{0}^{\infty} \frac{e^{-z(x+t)}}{1+x+t+xt} \, dx \, dt \\ &=2e^{-2z} \int_{0}^{\infty} \int_{0}^{t}\frac{e^{-z(x+t)}}{1+x+t+xt} \, dx \, dt \\ &= 2e^{-2z} \int_{0}^{\infty} \int_{0}^{u^{2}/4} \frac{e^{-zu}}{1+u+v} \frac{dv \, du}{\sqrt{u^{2}-4v}} \tag{1} \\ &= 4 e^{-2z} \int_{0}^{\infty} e^{-zu} \int_{0}^{u} \frac{1}{(2+u)^{2}-w^{2}} \, dw \, du \tag{2} \\ &= 4e^{-2z} \int_{0}^{\infty} e^{-zu} \, \frac{1}{2+u} \text{arctanh} \left( \frac{u}{2+u}\right) \, du \\ &= 2e^{-2z} \int_{0}^{\infty} e^{-zu} \, \frac{\ln(1+u)}{2+u} \, du \\ &= 2e^{-2z} \int_{0}^{\infty} e^{-2zy} \, \frac{\ln (1+2y)}{1+y} \, dy \tag{3} \end{align}$$

$(1)$ Make the change of variables $u = x+t, v=xt$.

$(2)$ Make the substitution $w^{2} = u^{2}-4v$.

$(3)$ Make the substitution $y = \frac{u}{2}$.

Solution 1:

Let the integral in question be \begin{align}\tag{1} I_{a} = \int_{0}^{\infty} e^{-st} \, \frac{\ln(1+a t)}{1 + t} \, dt. \end{align} For the case of $a=1$ the following is obtained.

By utilizing \begin{align} \int_{0}^{\infty} e^{-s t} \, \ln^{2} t \, dt = \frac{1}{s} \, ( \zeta(2) + (\gamma + \ln s)^{2} ) \end{align} for the change $t \to t+1$ leads to \begin{align}\tag{2} \int_{1}^{\infty} e^{-s t} \, \ln^{2}(t+1) \, dt = \frac{e^{s}}{s} \, ( \zeta(2) + (\gamma + \ln s)^{2} ) . \end{align} The integral \begin{align}\tag{3} \int_{0}^{1} e^{-st} \, \ln^{2}(t+1) \, dt &= e^{s} \left\{ 4 g(-2s) - 2 g(-s) + \frac{e^{-2s}}{s} \, (e^{2s} -1) \, \ln^{2} 2 - \frac{2 \, \ln 2}{s} (\gamma + \ln(2s) + \Gamma(0,2s)) \right\}, \end{align} where $g(x) = {}_{3}F_{3}(1,1,1; 2,2,2; x)$. Now, it is readily seen that \begin{align} I_{1} &= \int_{0}^{\infty} e^{-st} \, \frac{\ln(1+t)}{1+t} \, dt = \frac{s}{2} \, \int_{0}^{\infty} e^{-st} \, \ln^{2}(1+t) \, dt. \end{align} Making use of equations (2) and (3) the result \begin{align} \int_{0}^{\infty} e^{-st} \, \frac{\ln(1+t)}{1+t} \, dt &= e^{s} \, \left\{ 2 s \, g(-2s) - s \, g(-s) + \frac{1}{2} \, (1 - e^{-2s}) \, \ln^{2} 2 - \ln 2 \, (\gamma + \ln(2s) + \Gamma(0,2s)) \right. \\ & \hspace{15mm} \left. + \frac{1}{2} \, ( \zeta(2) + (\gamma + \ln s)^{2}) \right\} \end{align} where $g(x) = {}_{3}F_{3}(1,1,1; 2,2,2; x)$.

Solution 2:

Denote the Laplace transform of $\frac{\ln(1+at)}{1+bt}$ as $$I(s,a,b)=\int^\infty_0e^{-st}\ \frac{\ln(1+at)}{1+bt}\ dt$$ where $s,a,b\in\mathbb R^+$. (One may extend $s$ to the complex plane afterwards.) Differentiate under the integral sign to obtain the differential equation \begin{align} I(s,a,b)-bI_s(s,a,b) &=\int^\infty_0e^{-st}\ln(1+at)\ dt\\ &=\frac{a}{s}\int^\infty_0\frac{e^{-st}}{1+at}\ dt\\ &=\frac{e^{s/a}}{s}\int^\infty_\frac sa\frac{e^{-u}}{u}\ du\\ \end{align} Integrating back yields $$I(s,a,b)=\frac{e^{s/b}}{b}\int^s_\infty\frac{\exp\left(\frac{b-a}{ab}v\right)\operatorname{Ei}(-\frac{v}{a})}{v}\ dv\tag{*}$$

If $a=2b$, $(*)$ is reduced to $$I(s,2b,b)=\frac{e^{s/b}}{b}\int^s_\infty\frac{\exp\left(-\frac{v}{2b}\right)\operatorname{Ei}(-\frac{v}{2b})}{v}\ dv=\frac{e^{s/b}\left[\operatorname{Ei}(-\frac{s}{2b})\right]^2}{2b}$$ Setting $(s,a,b)=(2z,2,1)$ reproduces the identity stated in the question.

If $a=b$, $(*)$ becomes \begin{align} I(s,b,b) &=\frac{e^{s/b}}{b}\int^s_\infty\frac{\operatorname{Ei}(-\frac vb)}{v}\ dv\\ \end{align} but this integral can at best be expressed in terms of a generalised ${}_3F_3$ hypergeometric function. It is unlikely that nice closed forms can be obtained for $I(s,a,b)$ for other $a$ and $b$.