Dominant contribution from $\int_1^\infty x^a (x-1)^b e^{cx}\ dx$

Question

For real $a\gg0$, $b\gg0$, $c<0$, I want to know if the dominant contribution from the integral, $$\int_1^\infty x^a (x-1)^b e^{cx}\ dx=\int_1^\infty\exp\left(a\log x+b\log(x-1)+cx\right)\ dx$$ is $$\int_{T-\epsilon}^{T+\epsilon}x^a(x-1)^be^{cx}\ dx=\int_{T-\epsilon}^{T+\epsilon}\exp\left(a\log x+b\log(x-1)+cx\right)\ dx$$ where $T$ solves the derivative of the exponent, $$\frac{a}{x}+\frac{b}{x-1}+c=0.$$

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I assumed this was true and tried building an asymptotic approximation for large $T$.

Let $u=x-T$, $$\int_{-\epsilon}^\epsilon\exp\left(a\log(u+T)+b\log(u+T-1)+c(u+T)\right)\ du$$ note that when $u$ is close to $0$, $$a\log(u+T)+b\log(u+T-1)+c(u+T)\sim a\log(T)+b\log(T-1)+cT+\left(\frac{a}{T}+\frac{b}{T-1}+c\right)u+O\left(u^2\right)$$ keeping the first non-constant term and expanding the terms beyond it by its Maclaurin series, $$\int_{-\epsilon}^\epsilon\exp \left(a\log(T)+b\log(T-1)+cT+\left(\frac{a}{T}+\frac{b}{T-1}+c\right)u\right)\left(1+O\left(u^2\right)\right)\ du$$ hence the leading approximation is, $$\exp\left(a\log(T)+b\log(T-1)+cT\right)\int_{-\epsilon}^\epsilon\exp\left(\left(\frac{a}{T}+\frac{b}{T-1}+c\right)u\right)\ du$$ normally at this step I would replace the end points of the integral with $-\infty$ and $\infty$, but here I get a divergent result. Does this mean that the terms I'm introducing are not subdominant, i.e., my initial assumption of rewriting as $\int_{T-\epsilon}^{T+\epsilon}$ is incorrect?

Answer 1

I believe I found the issue. The coefficient of the $u^1$ term was $0$ by the definition of $T$, so taking one more term should resolve the issue. $$a\log(T)+b\log(T-1)+cT-\left(\frac{a}{T^2}+\frac{b}{(T-1)^2}\right)\frac{u^2}{2}+O\left(u^3\right)$$ this time properly keeping the first non-constant term, $$\exp\left(a\log(T)+b\log(T-1)+cT\right)\int_{-\epsilon}^{\epsilon}\exp{\left(-\left(\frac{a}{T^2}+\frac{b}{(T-1)^2}\right)\frac{u^2}{2}\right)}\left(1+O\left(u^3\right)\right)\ du$$ hence the leading order, $$\exp\left(a\log(T)+b\log(T-1)+cT\right)\int_{-\infty}^{\infty}\exp{\left(-\left(\frac{a}{T^2}+\frac{b}{(T-1)^2}\right)\frac{u^2}{2}\right)}\ du$$ where we also replaced the endpoints of the integral. Rewrite the integrand as $\exp(-ku^2/2)$ and note that it is even, $$2\exp\left(a\log(T)+b\log(T-1)+cT\right)\int_0^\infty\exp\left(-\frac{ku^2}{2}\right)\ du$$ with $$k:=\frac{a}{T^2}+\frac{b}{(T-1)^2}$$ perform a change of variables from $u$ to $s$ via $s^2=ku^2/2$, $$2\sqrt{\frac{2}{k}}\exp\left(a\log(T)+b\log(T-1)+cT\right)\int_0^\infty\exp \left(-s^2\right)\ ds$$ hence by the Gaussian integral, $$\begin{align*}\int_1^\infty x^a (x-1)^b e^{cx}\ dx&\sim\sqrt{\frac{2\pi}{k}}\exp\left(a\log(T)+b\log(T-1)+cT\right) \\&=\sqrt{\frac{2\pi}{k}}T^a(T-1)^be^{cT}\end{align*}$$ as $T\to\infty$. Some simplifications with $k$ may be possible.

Answer 2

Just a few ideas.

May be, you could consider the two (limiting ?) cases $$I_1=\int_1^\infty x^a\, x^b \, e^{cx},\,dx=E_{-(a+b)}(-c)$$ $$I_2=\int_1^\infty (x-1)^a\, (x-1)^b \, e^{cx}\,dx= \frac{e^c\,\,\Gamma (a+b+1)}{(-c)^{(a+b+1)}}$$ or, even, for an appropriate value of $k \in (0,1)$ $$I_3=\int_1^\infty (x-k)^a\, (x-k)^b \, e^{cx}\,dx=\frac{ e^{c k} \,\,\Gamma (a+b+1,-c (1-k))}{(-c)^{(a+b+1)}}$$ which simplifies the problem from $3$D to $2$D.

Also, notice that $$I_4=\int_1^\infty x^\alpha\, (x-1)^\alpha \, e^{cx}\,dx=\frac{e^{c/2}\,\Gamma(\alpha+1)}{(-c)^{\frac{2\alpha+1}{2}}\sqrt{\pi }}K_{\frac{2\alpha+1}{2}}\left(-\frac{c}{2}\right)$$

Now, consider the behaviour of $(I_1,I_2,I_3,I_4)$ on a logarithmic scale.