Consider the following integral $$ \int_1^\infty x^{k-b} e^{-ck x} dx $$ where $b>0$ and $c\in(0,1)$ are fixed and $k \to \infty$. What is the behaviour of this integral as $k$ diverges?
I have tried to re-express it as $(ck)^{a+1-k}\Gamma(k-a,ck)$, but this has not been very helpful as I haven't managed to find a suitable characterization of the limiting behaviour of $\Gamma(k-a,ck)$. If anyone could point me to a useful reference, it would be very helpful.
On
Assume that $b,c>0$. Svyatoslav's asymptotics is valid only if $c$ is bounded away from $1$ and $0<c<1$. I shall obtain an asymptotic approximation for large $k$, which is uniformly valid for all $c>0$, using Bleistein's method. First, we perform some changes of integration variables: \begin{align*} I_{b,c} (k) = \int_1^{ + \infty } {x^{k - b} {\rm e}^{ - ckx} dx} &= ({\rm e}c)^{ - k} \int_0^{ + \infty } {{\rm e}^{ - k(c{\rm e}^t - t - 1 - \log c)} {\rm e}^{(1 - b)t} {\rm d}t} \\ & = ({\rm e}c)^{ - k} \int_0^{ + \infty } {\exp\! \left( { - \tfrac{k}{2}(w - C)^2 } \right){\rm e}^{(1 - b)t} \frac{{{\rm d}t}}{{{\rm d}w}}{\rm d}w} . \end{align*} Here $$ C = \sqrt {2(c - 1 - \log c)}, $$ and $$ c{\rm e}^t - t - 1 - \log c = \tfrac{1}{2}(w - C)^2 . $$ The branch of the square root is chosen so that $C \sim 1 - c$ as $c\to 1$. Now, as $w\to C$, $$ {\rm e}^{(1 - b)t} \frac{{{\rm d}t}}{{{\rm d}w}} = c^{b - 1} + \left( {\tfrac{2}{3} - b} \right)c^{b - 1} (w - C) + \mathcal{O}((w - C)^2 ). $$ Hence, \begin{align*} I_{b,c} (k) & \sim c^{b - 1} ({\rm e}c)^{ - k} \int_{ - C}^{ + \infty } {\exp\! \left( { - \tfrac{k}{2}s^2 } \right){\rm d}s} + \left( {\tfrac{2}{3} - b} \right)c^{b - 1} ({\rm e}c)^{ - k} \int_{ - C}^{ + \infty } {\exp\! \left( { - \tfrac{k}{2}s^2 } \right)s\,{\rm d}s} \\ & = c^{b - 1} ({\rm e}c)^{ - k} \sqrt {\frac{\pi }{{2k}}} {\mathop{\rm erfc}\nolimits} \!\left( { - C\sqrt {\tfrac{k}{2}} } \right) + \left( {\tfrac{2}{3} - b} \right)c^{b - 1} ({\rm e}c)^{ - k} \frac{1}{k}\exp \!\left( { - C^2 \tfrac{k}{2}} \right) \end{align*} as $k\to+\infty$. Here $\text{erfc}$ is the complementary error function. In summary, $$ I_{b,c} (k) \sim c^{b - 1} ({\rm e}c)^{ - k} \sqrt {\frac{\pi }{{2k}}} \left( {{\mathop{\rm erfc}\nolimits} \!\left( { - C\sqrt {\tfrac{k}{2}} } \right) + \left( {\tfrac{2}{3} - b} \right)\sqrt {\frac{2}{{\pi k}}} \exp\! \left( { - C^2 \tfrac{k}{2}} \right)} \right) $$ as $k\to+\infty$, uniformly in $c>0$ and with any fixed $b>0$.
If $0<c<1$ and $c$ is bounded away from $1$, the asymptotics may be simplified to $$ I_{b,c} (k) \sim c^{b - 1} ({\rm e}c)^{ - k} \sqrt {\frac{{2\pi }}{k}} $$ using $(7.12.1)$, which is in agreement with the result by Svyatoslav.
Method $1$
The Laplace' method can be applied here to find the asymptotics at $k\to\infty$.
We suppose that $0<c<1$. Then
$$I(k, b,c)=\int_1^\infty x^{k-b}e^{-ckx}dx=\int_1^\infty x^{-b}e^{-k(cx-ln x)}dx=\int_1^\infty x^{-b}e^{-kf(x)}dx$$ $$f(x)=cx-\ln x\,\Rightarrow\, f'(x_0)=0\,\,\text{at}\,\,x_0=\frac1c$$ $$f(x)=f(x_0)+\frac12f''(x_0)(x-x_0)^2+...\,=1+\ln c+\frac{c^2(x-\frac1c)^2}2+...$$ and $$I\sim\int_1^\infty x^{-b}e^{-k\big(1+\ln c+\frac{c^2}2(x-\frac1c)^2\big)}dx\sim(c\,e)^{-k}c^b\int_{-\infty}^\infty e^{-\frac {kc^2}2t^2}dt$$ $$\boxed{\,\,I(k,b,c)\sim\sqrt\frac{2\pi}kc^{b-1}(c\,e)^{-k}\,\,}$$ As @Claude Leibovici suggested, we can try to find another approach.
Method $2$
An approach that begs for itself is to extend the integration to $0$ and use the asymptotics of gamma-function. But we have to prove that $$\lim_{k\to\infty}\frac{\int_0^1x^{k-b}e^{-ckx}dx}{\int_0^\infty x^{k-b}e^{-ckx}dx}=\lim_{k\to\infty}\frac{I_2(k)}{I_1(k)}=0$$ First, $$I_1=\int_0^\infty x^{k-b}e^{-ckx}dx=\frac1{(kc)^{k-b+1}}\int_0^\infty t^{k-b}e^{-t}dt=\frac{\Gamma(k-b+1)}{(kc)^{k-b+1}}$$ Using the Stirling's formula ($k\to\infty$) $$I_1\sim\frac{\sqrt{2\pi(k-b)}\left(\frac{k-b+1}e\right)^{k-b+1}}{(kc)^{k-b+1}}\sim\sqrt\frac{2\pi}k\frac{e^{b-k-1}}{c^{k-b+1}}e^{(k-b+1)\ln(1-\frac{b-1}k)}$$ $$I_1\sim\sqrt\frac{2\pi}kc^{b-1}(c\,e)^{-k}\tag{1}$$ Now, for $I_2(k)$ $$I_2(k)=\int_0^1x^{k-b}e^{-ckx}dx=\int_0^1(1-t)^{k-b}e^{-ck(1-t)}dt$$ $$=e^{-ck}\int_0^1e^{ckt+(k-b)\ln(1-t)}dt<e^{-ck}\int_0^1e^{ckt-(k-b)t}dt$$ $$I_2<\frac{e^{-ck}e^b}{k(1-c)}\tag{2}$$ Comparing (1) and (2), it is enough to prove that $$\lim_{k\to\infty}\frac{e^{-kc}}{(ce)^{-k}}=0, \,\,\text{or}\,\,e^c>c\,e$$ This in fact is valid for $c\in(0;1)$ - both functions are growing, equal at $c=1$, and $\,e^c>c\,e\,$ at $\,c=0$.
Hence, $$\lim_{k\to\infty}\frac{I_2(k)}{I_1(k)}=0$$ and $$\boxed{\,\,I(k,b,c)\sim I_1(k)\sim \sqrt\frac{2\pi}kc^{b-1}(c\,e)^{-k}\,\,}$$