Can we use Laplace's method to obtain the large x behaviour of $\int_0^\infty e^{-x(t^3+s^2+s)} ds$?

Question

How to evaluate $\int_0^\infty e^{-x(t^3+s^2+s)} ds$?

I know the solutions to $\int_0^\infty e^{-x(t^3)} ds$ ,$\int_0^\infty e^{-x(s^2)} ds$ ,and$\int_0^\infty e^{-x(s)} ds$ seperately, but when they are multiplied together, what can I do?

Since x is involed, I believe it is not correct to neglect any term in the upper script. What are the general procedures for evaluating this type of integral?

Answer 1

Some quite lenient bounds can be obtained the following way: Since:

$$ \left(t+\frac{1}{3}\right)^{3}-\frac{1}{27}<t^{3}+t^{2}+t<\left(t+\frac{1}{\sqrt{3}}\right)^{3}-\frac{1}{\sqrt{27}} $$ And: $$ x^{-1/3}\Gamma\left(\frac{4}{3}\right)=\int_{0}^{\infty}\exp\left(-xt^{3}\right)\mathrm{d}t $$ Hence:

$$ \int_{0}^{\infty}\exp\left(-x\left(t+t^{2}+t^{3}\right)\right)\mathrm{d}x\leq\exp\left(-\frac{x}{27}\right)\int_{0}^{\infty}\exp\left(-x\left(t+\frac{1}{3}\right)^{3}\right)\mathrm{d}t= $$

$$ =\exp\left(-\frac{x}{27}\right)\left(x^{-1/3}\Gamma\left(\frac{4}{3}\right)-\int_{0}^{1/3}\exp\left(-xt^{3}\right)\mathrm{d}t\right) $$

$$ \int_{0}^{\infty}\exp\left(-x\left(t+t^{2}+t^{3}\right)\right)\mathrm{d}x\geq\exp\left(-\frac{x}{\sqrt{27}}\right)\int_{0}^{\infty}\exp\left(-x\left(t+\frac{1}{\sqrt{3}}\right)^{3}\right)\mathrm{d}t= $$

$$ =\exp\left(-\frac{x}{\sqrt{27}}\right)\left(x^{-1/3}\Gamma\left(\frac{4}{3}\right)-\int_{0}^{1/\sqrt{3}}\exp\left(-xt^{3}\right)\mathrm{d}t\right) $$

$$ \int_{0}^{1/\sqrt{3}}\exp\left(-xt^{3}\right)\mathrm{d}t=\frac{\Gamma \left(\frac{4}{3}\right)}{\sqrt[3]{x}}-\frac{\alpha \Gamma \left(\frac{1}{3},x \alpha ^3\right)}{3 \sqrt[3]{\alpha ^3 x}} $$ Where $\Gamma(x,y)$ is the incomplete gamma function. At $\infty$, we have the following expansion: $$ \frac{\alpha \Gamma \left(\frac{1}{3},x \alpha ^3\right)}{3 \sqrt[3]{\alpha ^3 x}}=e^{\alpha ^3 (-x)} \left(\frac{1}{3 \alpha ^2 x}-\frac{2}{9 \alpha ^5 x^2}+\frac{10}{27 \alpha ^8 x^3}+O\left(\left(\frac{1}{x}\right)^{11/3}\right)\right) $$

Answer 2

OP's integral (v5) reads

$$I(x,t)~:=~ \int_{\mathbb{R}_+}\!ds~ e^{-x(t^3+s^2+s)} ~=~ e^{x(1/4-t^3)} J_+(x) ,\tag{1}$$ where $$J_{\pm}(x)~:=~ \int_{\mathbb{R}_+}\!ds~ e^{-x(s\pm 1/2)^2}.\tag{2}$$ Note that the sum is a Gaussian integral $$J_+(x)+J_-(x)~=~ \sqrt{\frac{\pi}{x}} .\tag{3}$$ The other integral $$ J_-(x)~\approx~\sqrt{\frac{\pi}{x}} e^{-x/4}\quad\text{for}\quad x~\to \infty\tag{4} $$ can be evaluate via Laplace's method, as OP wanted to know. Putting everything together, we get $$I(x,t)~\approx~e^{-xt^3}\sqrt{\frac{\pi}{x}} (e^{x/4}-1)\quad\text{for}\quad x~\to \infty.\tag{5} $$