Asymptotic behaviour/upper bound for $\int_0^{\infty} \exp(-c x^a+K x^b)dx$ for $a>b>0$ as $K\rightarrow \infty$?

What is the asymptotic behaviour or an upper bound for $\int_0^{\infty} \exp(-c x^a+K x^b) \, dx$, for $a>b>0,$ as $K\rightarrow \infty$? Or any good reference for tools to tackle this question?

I think the growth in $K$ should be polynomial because $-c x^a+K x^b=0$ yields $x=(K/C)^\frac{1}{a-b}$ on the range $[0,(K/C)^\frac{1}{a-b}]$ the maximum value of the integrand is again a power of K (take derivative and set 0) the product yields an upper bound on $\int_0^{(K/C)^\frac{1}{a-b}} \exp(-c x^a+K x^b) \, dx$.

On the other hand $\int_{(K/C)^\frac{1}{a-b}}^{\infty} \exp(-c x^a+K x^b) \, dx$ should be decreasing in $K$, for large $K$.

Thank you,

One can get a full asymptotic expansion as an application of Watson's Lemma. One need only observe that the integrand is maximized at $x_0 = \left(\frac{Kb}{ac}\right)^{1/(a-b)}.$
Substituting $x = x_0 u,$ one gets(where $I$ is the original integral) $I = x_0 \int_0^\infty \exp(-c^{b/(a-b)} K^{a/(a-b)}((b/a)^{a/(a-b)} u^a - (b/a)^{b/(a-b)} u^b)) d u.$ Letting $t = c^{b/(a-b)} K^{a/(a-b)},$ the integral breaks up into two Watson Lemma integrals, one from $0$ to $1,$ the second from $1$ to $\infty.$ I leave the final computation of the asymptotics to the interested reader.