Evaluate integral of $\ln(u)\exp\{-bu\}/u du$

Question

While doing my research, I came across this integral and don't know how to solve for this: $$\int_{0}^{\infty}x^2\exp\{ax-be^{ax}\}dx,\text{where $a,b>0$}.$$ My attempt: \begin{align} \int_{0}^{\infty}x^2\exp\{ax-be^{ax}\}dx &\overset{x = \ln u}{=} \int_1^{\infty}\frac{1}{u}\left(\ln u\right)^{2}\exp\left\{a\ln u-be^{a \ln u}\right\}du\\ &=\int_{1}^{\infty}(\ln u)^2u^{a-1}e^{-bu^a}du\\ &\overset{t=u^a}{\propto} \int_{1}^{\infty}(\ln t)^2e^{-bt}dt = A \end{align} Then, integration by part where $k=(\ln t)^2$, we have: $$A\propto \int_{1}^{\infty}\frac{1}{t}(\ln t )\exp\{-bt\}dt=B$$ Here, my first approach is to use Taylor Series expansion and obtain: $$B=\int_{1}^{\infty}\frac{1}{t}(\ln t)\sum_{n=0}^\infty\frac{(-bt)^n}{n!}dt=\int_{1}^{\infty}\sum_{n=0}^{\infty}(-1)^n\frac{b^nt^{n-1}\ln t}{n!}dt$$ However, here $f_{n}(t) = (-1)^n\frac{b^nt^{n-1}\ln t}{n!}$ is not greater than $0$ for all values of $t$ since $b > 0$, so I can't interchange integration and summation. (by Fubini's theorem)

Then, my second approach I integration by part one more time: $$B=\int_{1}^{\infty}\frac{1}{t}(\ln t )\exp\{-bt\}dt\overset{k=\ln x}{=}\int_{0}^{\infty}k\exp\{-be^k\}dk$$ Let $y=k$ and $dz=\exp\{-be^k\}dk \rightarrow z = -E_1\left(be^k\right)$ So, $$B=-tE_1\left(be^k\right)\big|_{0}^{\infty}+\int_{0}^{\infty}E_1\left(be^k\right)dk$$ where $E_1\left(be^k\right)$ is the exponential integral of $be^k$.

But now, I don't know where to go from here. Any suggestion?

Answer 1

Starting with $A$, and knowing that the constant of proportionality is $a^{-3}$, we can see that \begin{align} A&=a^{-3}\int_1^\infty \log^2 x\, e^{-bx}\,dx \\ &= a^{-3}\frac{\partial^2}{\partial \mu^2} \int_1^\infty x^{\mu-1} e^{-bx}\,dx \Big|_{\mu=1} \\ &= a^{-3}\frac{\partial^2}{\partial \mu^2} b^{-\mu} \int_b^\infty z^{\mu-1} e^{-z}\,dz \Big|_{\mu=1} \\ &= a^{-3}\frac{\partial^2}{\partial \mu^2} b^{-\mu} \,\Gamma(\mu, b) \Big|_{\mu=1}\\ \end{align} Where $\Gamma(\mu, b)$ is the upper incomplete gamma function. Now, if you want to go further, we can use the lower incomplete gamma function and its series representation:

\begin{align} A&= a^{-3}\frac{\partial^2}{\partial \mu^2} b^{-\mu} \Big\{ \left( \Gamma(\mu)-\gamma(\mu, b)\right)\Big\} \Big|_{\mu=1}\\ &= a^{-3}\frac{\partial^2}{\partial \mu^2} \Big\{ b^{-\mu} \Gamma(\mu)-b^{-\mu} \gamma(\mu, b) \Big\} \Big|_{\mu=1}\\ &= a^{-3}\frac{\partial^2}{\partial \mu^2}\Big\{ b^{-\mu} \Gamma(\mu)-b^{-\mu} \sum_{n\ge 0} \frac{(-1)^n}{n!} \frac{b^{\mu+n}}{\mu+n} \Big\}\Big|_{\mu=1}\\ &= a^{-3}\frac{\partial^2}{\partial \mu^2} \Big\{b^{-\mu} \Gamma(\mu)- \sum_{n\ge 0} \frac{(-1)^n}{n!} \frac{b^n}{\mu+n} \Big\} \Big|_{\mu=1}\\ &= a^{-3}\frac{\log^2 b +2\gamma\log b +\pi^2/6 + \gamma^2}{b}- 2a^{-3}\sum_{n\ge 0} \frac{(-1)^n}{n!} \frac{b^n}{(n+1)^3}\\ \end{align}

Answer 2

The integral you call $A$ can be split into two integrals.

$$ A = \frac{1}{a^{3}}\int_{1}^{\infty}\ln^{2}t\,e^{-bt}\,\mathrm{d}t = \frac{1}{a^{3}}\left[\int_{0}^{\infty}\ln^{2}t\,e^{-bt}\,\mathrm{d}t - \int_{0}^{1}\ln^{2}t\, e^{-bt}\,\mathrm{d}t\right] = \frac{1}{a^{3}}(A_{1} - A_{2}) $$

The first one can be done by considering the series expansion

$$\begin{aligned} \int_{0}^{\infty}t^{\epsilon}e^{-bt}\,\mathrm{d}t &= \sum_{n=0}^{\infty}\frac{\epsilon^{n}}{n!}\int_{0}^{\infty}\ln^{n}t\,e^{-bt}\,\mathrm{d}t \\ &= \frac{1}{b^{1+\epsilon}}\int_{0}^{\infty}t^{\epsilon}e^{-t}\,\mathrm{d}t = \frac{\Gamma(1+\epsilon)}{b^{1+\epsilon}}\end{aligned}$$

and expanding the latter term to second order in $\epsilon$. The integral will then be $2! = 2$ times the coefficient of the $\epsilon^{2}$ term to compensate for the factorial. Using the Taylor expansion

$$ \ln\Gamma(1+\epsilon) = -\gamma\epsilon + \sum_{k=2}^{\infty}\frac{(-1)^{k}\zeta(k)}{k}\epsilon^{k},$$

we have

$$\begin{aligned} \frac{\Gamma(1+\epsilon)}{b^{1+\epsilon}} &\approx \frac{e^{-\gamma\epsilon + \frac{\zeta(2)}{2}\epsilon^{2}}}{be^{\epsilon\ln b}} \approx \frac{1}{b}\left(1 - \epsilon\ln b + \frac{\ln^{2}b}{2}\epsilon^{2}\right)\left(1 - \gamma\epsilon + \frac{\zeta(2)}{2}\epsilon^{2} + \frac{\gamma^{2}}{2}\epsilon^{2}\right) \\ &\approx \frac{1}{b}\left(1 + \left(-\gamma - \ln b\right)\epsilon +\left(\frac{\zeta(2)}{2} + \frac{\gamma^{2}}{2} + \gamma\ln b + \frac{\ln^{2}b}{2}\right)\epsilon^{2}\right)\end{aligned}$$

so

$$ A_{1} = \frac{1}{b}\left(\zeta(2) + \gamma^{2} + 2\gamma\ln b + \ln^{2}b\right). $$

The second integral can be done using series again, but I am not sure if it has a nice expression. I write it in terms of hypergeometric functions. The Taylor series of the exponential is used. (Note I can switch sum and integral because $\int_{0}^{1}t^{k}\ln^{2}t\,\mathrm{d}t$ certainly converges for any whole number $k$, as the divergence of $\ln^{2}t$ is weak)

$$ A_{2} = \int_{0}^{1}\ln^{2}t\,\sum_{k=0}^{\infty}\frac{(-1)^{k}b^{k}t^{k}}{k!}\,\mathrm{d}t = \sum_{k=0}^{\infty}\frac{(-1)^{k}b^{k}}{k!}\int_{0}^{1}t^{k}\ln^{2}t\,\mathrm{d}t $$

Consider

$$ \int_{0}^{1}t^{k+\epsilon}\,\mathrm{d}t = \sum_{n=0}^{\infty}\frac{\epsilon^{n}}{n!}\int_{0}^{1}t^{k}\ln^{n}t\,\mathrm{d}t = \frac{1}{k+\epsilon+1}$$

where we are again finding the $\epsilon^{2}$ coefficient. It follows that

$$ \frac{1}{1+k+\epsilon} = \frac{1}{1+k}\frac{1}{1+\frac{\epsilon}{1+k}} = \frac{1}{1+k}\sum_{j=0}^{\infty}(-1)^{j}\left(\frac{\epsilon}{1+k}\right)^{j}$$

so

$$ \int_{0}^{1}t^{k}\ln^{2}t\,\mathrm{d}t = \frac{2}{(1+k)^{3}} \quad\to\quad A_{2} = \sum_{k=0}^{\infty}\frac{(-1)^{k}b^{k}}{k!}\frac{2}{(1+k)^{3}} = 2\,{}_{3}F_{3}(1,1,1;2,2,2;-b).$$