Calculate the integrals $\int\limits_{0}^{\infty }e^{-\lambda (x-1)^2}dx+\int\limits_{0}^{\infty }e^{-\lambda (x-2)^2}dx$

Question

In the $\lambda \to \infty $ limit, approximate the integral $$I(\lambda )=\int\limits_{0}^{\infty }e^{-\lambda (x-1)^2(x-2)^2}dx$$

I understand that the function $-\lambda (x-1)^2(x-2)^2$ reaches a minimum at the points $x = 1$ and $x = 2$, so you can decompose the function $$x=1\Rightarrow -\lambda (x-1)^2(x-2)^2 \approx -\lambda (x-1)^2$$ $$x=2\Rightarrow -\lambda (x-1)^2(x-2)^2 \approx -\lambda (x-2)^2$$

$$I(\lambda )=\int\limits_{0}^{\infty }e^{-\lambda (x-1)^2(x-2)^2}dx\approx \int\limits_{0}^{\infty }e^{-\lambda (x-1)^2}dx+\int\limits_{0}^{\infty }e^{-\lambda (x-2)^2}dx$$

How do I calculate these integrals? As I understand it, they are not expressed in terms of elementary functions

If $\lambda$ approaches $\infty$, wouldn't the integrand approach $0$? — Kamal Saleh, May 22 '23 at 15:31
I don’t know whether $I(\lambda)$ can be aproximated as the sum of 2 integrals like that — NN2, May 22 '23 at 20:53

score 2 · Accepted Answer · answered May 22 '23 at 16:40

These integrals can be computed directly:

For $a>0$

$$I_a(\lambda)=\int^\infty_0e^{-\lambda(x-a)^2}\,dx=\frac{1}{\sqrt{\lambda}}\int^\infty_{-a\sqrt{\lambda}}e^{-u^2}\,du $$ Thus $$\sqrt{\lambda}\int^\infty_0 e^{-\lambda(x-a)^2}\,dx=\int^\infty_{-a\sqrt{\lambda}}e^{-u^2}\,du\xrightarrow{\lambda\rightarrow\infty}\sqrt{\pi}$$

That is

$$I_1(\lambda)+I_2(\lambda)\sim 2\sqrt{\frac{\pi}{\lambda}}\quad\text{as}\quad\lambda\rightarrow\infty$$

Calculate the integrals $\int\limits_{0}^{\infty }e^{-\lambda (x-1)^2}dx+\int\limits_{0}^{\infty }e^{-\lambda (x-2)^2}dx$

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