0

For all $\lambda>0,\qquad f(\lambda)=\displaystyle \int_0^{+\infty} \dfrac{e^{-\lambda t^2}}{1+t^2} \cdot dt$

  1. Solve $f(0)$

  2. Show that $f(\lambda)=\mathcal{O}_{+\infty}\left(\dfrac{1}{\sqrt{\lambda}}\right)$

My attempt :

  1. $f(0)=\displaystyle \int_0^{+\infty} \dfrac{1}{1+t^2} \cdot dt=\lim_{x\to +\infty} \arctan(x)=\dfrac{\pi}{2}$ $\\$

  2. $f(\lambda)=\mathcal{O}\left(\dfrac{1}{\sqrt{\lambda}}\right)\iff\exists C>0,\;\exists A\ge0,\; \forall \lambda>A,\qquad \sqrt{\lambda}f(\lambda)\le C$

So $$\displaystyle \sqrt{\lambda}f(\lambda)= \int_0^{+\infty} \dfrac{\sqrt{\lambda}}{1+t^2} \cdot e^{-\lambda t^2}\cdot dt$$

Now let $g(t):=\dfrac{\sqrt{\lambda}}{1+t^2}$ and $f(t):=e^{-\lambda t^2}$

for any $\lambda>0$ we have :

$\left\lbrace\begin{array}l \displaystyle \int^{+\infty}_0 g(t) \cdot dt \quad \text{converge}\\\\f(t)>0\\ f\; \text{is decreasing} \end{array}\right.\quad $ so the "Abel criterion" guarantiees $\sqrt{\lambda}f(\lambda)$ converges

Thus $\forall \lambda >0, \exists \ell>0$ such that $\sqrt{\lambda}f(\lambda)=\ell$

Conclusion and question :

I know that $\sqrt{\lambda}f(\lambda)$ is decreasing but I don't know how to prove it. If I assume $\sqrt{\lambda}f(\lambda)$ is decreasing, I choose $C=\dfrac{\pi}{2}=f(0)$ and the proof is complete, we have $f(\lambda)=\mathcal{O}\left(\dfrac{1}{\sqrt{\lambda}}\right)$

Stu
  • 1,690
  • 1
    It requires more work than you should actually do here, but in fact you can get a more precise asymptotic for this integral using the Laplace method. – Ian Oct 04 '17 at 17:30

2 Answers2

1

hint

If we put $$u=t\sqrt {\lambda}$$

we get $$\sqrt {\lambda}f (\lambda)\le \int_0^{+\infty}e^{-u^2}du=C $$

using that $$\frac {1}{1+t^2}\le 1.$$

hamam_Abdallah
  • 62,951
1

Hint

$$ \int_0^{x}\frac{\sqrt{\lambda}}{1+t^2}e^{-\lambda t^2}dt\le\int_0^{x}\sqrt{\lambda} e^{-\lambda t^2}dt$$

Then let $u=\sqrt{\lambda} t$