Krdytkyu

While doing my research, I came across this integral and don't know how to solve for this:
$$int_{0}^{infty}x^2exp{ax-be^{ax}}dx,text{where $a,b>0$}.$$
My attempt:
begin{align}
int_{0}^{infty}x^2exp{ax-be^{ax}}dx &overset{x = ln u}{=} int_1^{infty}frac{1}{u}left(ln uright)^{2}expleft{aln 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:
$$Apropto int_{1}^{infty}frac{1}{t}(ln t )exp{-bt}dt=B$$
Here, my first approach is to use Taylor Series expansion and obtain:
$$B=int_{1}^{infty}frac{1}{t}(ln t)sum_{n=0}^inftyfrac{(-bt)^n}{n!}dt=int_{1}^{infty}sum_{n=0}^{infty}(-1)^nfrac{b^nt^{n-1}ln t}{n!}dt$$
However, here $f_{n}(t) = (-1)^nfrac{b^nt^{n-1}ln t}{n!}$ 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:
$$B=int_{1}^{infty}frac{1}{t}(ln t )exp{-bt}dtoverset{k=ln x}{=}int_{0}^{infty}kexp{-be^k}dk$$
Let $y=k$ and $dz=exp{-be^k}dk rightarrow z = -E_1left(be^kright)$
So,
$$B=-tE_1left(be^kright)big|_{0}^{infty}+int_{0}^{infty}E_1left(be^kright)dk$$
where $E_1left(be^kright)$ is the exponential integral of $be^k$.

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

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_{nge 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_{nge 0} frac{(-1)^n}{n!} frac{b^n}{mu+n} Big} Big|_{mu=1}\
&= a^{-3}frac{log^2 b +2gammalog b +pi^2/6 + gamma^2}{b}- 2a^{-3}sum_{nge 0} frac{(-1)^n}{n!} frac{b^n}{(n+1)^3}\
end{align}

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

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

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

$$ lnGamma(1+epsilon) = -gammaepsilon + sum_{k=2}^{infty}frac{(-1)^{k}zeta(k)}{k}epsilon^{k},$$

we have

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

so

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

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

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