3

the main goal is to compute it and I think the intended way to do that is using Feynman's trick. meaning we will have to somehow differentiate things.

so how do you show that the function in the title is differentiable ?

I don't really have an idea how to approach the study of differentiablity in this case, my first guess would be that it's differentiable because since :

A) $\lambda \mapsto e^{-x^2} \cos {\lambda x} $ is infinitely differentiable

B) $|e^{-x^2} \cos {\lambda x}| \leq e^{-x^2} \in L^1(\mathbb{R})$

then the dominated convergence theorem guarantees the validity of swapping limits with integral signs, right ? so is that enough/correct ?

now assuming I can differentiate under the integral sign, if I do it twice I obtain the following ODE :

$$\Lambda(\lambda) +\lambda^2\Lambda''(\lambda) = 0 $$

which frankly looks quite like a Cauchy-Euler equation so plugging in $\Lambda(\lambda) = \lambda ^{\mu}$

we get $ \mu^2 -\mu +1 = 0 \iff (\mu - \frac12)^2 = -\frac34 \iff \mu =\frac12 \pm i\frac{\sqrt 3}{2}$

so if I have a good memory the solution should be $$ \lambda =\sqrt \lambda( A \cos {\frac{\sqrt 3}{2} \ln \lambda} + B\sin {\frac{\sqrt 3}{2} \ln \lambda} )$$

I don't wanna go further because normally $\Lambda(0) = \sqrt{\pi}$,

anything that can help me tackle this problem will be greatly appreciated.

also please confirm if my way of proving differentiability is correct or not, thanks !

Edit : I'm dumb the ODE should be : $$\Lambda(\lambda) +x^2\Lambda''(\lambda) = 0 $$

will work on it and update.

Edit 2 : I'm $\text{dumb}^2$

you can't take the $x$ outside of the integral of course ! will update.

the_firehawk
  • 2,425
  • Fix the solution hould be $$\color{red}{\Lambda} =\sqrt \lambda( A \cos {\frac{\sqrt 3}{2} \ln \lambda} + B\sin {\frac{\sqrt 3}{2} \ln \lambda} )$$ – Claude Leibovici May 16 '18 at 07:47
  • 1
    @ClaudeLeibovici yeah I realized that right after the post submission, thanks – the_firehawk May 16 '18 at 07:49
  • 1
    Concerning differentiability of integrals: https://math.stackexchange.com/q/519176/8157 (This is the most general theorem I have seen to justify this kind of operation) – Giuseppe Negro May 16 '18 at 08:54

2 Answers2

4

This might work: $\Lambda'(\lambda)=-\int_{\mathbb R} e^{-x^2}x\sin(\lambda x)dx=-\frac{1}{2}\lambda \Lambda(\lambda)$ after an integration by parts.

user254433
  • 2,733
2

Notice $\Lambda(\lambda)$ looks like a Fourier transform of $e^{-x^2}$

$\Lambda(\lambda)=\displaystyle\int_{-\infty}^\infty e^{-x^2}\cos(\lambda x)dx$

$\operatorname{Re}(\displaystyle e^{-i\omega t})=\cos(\omega t)\longrightarrow\operatorname{Re}(e^{-i\lambda x})=\cos(\lambda x)$

So

$$\displaystyle\Lambda(\lambda)=\int_{-\infty}^\infty e^{-x^2}\operatorname{Re}\left(e^{-i\lambda x}\right)dx=\int_{-\infty}^\infty\operatorname{Re}\left(e^{-x^2}e^{-i\lambda x}\right)dx=\operatorname{Re}\left(\int_{-\infty}^\infty e^{-x^2}e^{-i\lambda x}dx\right)$$

$$\Lambda(\lambda)=\operatorname{Re}\left(\mathcal{F}\left(e^{-x^2}\right)\right)=\mathcal{F}\left(e^{-x^2}\right)$$

I'm referring to this post at this point to transform

Denote $f(x)=e^{-x^2}$

$$f'(x)=-2xf(x)$$

$$-2i\frac{d}{d\xi}\hat{f}(\xi)=i\xi\hat{f}(\xi)\\\hat{f}'(\xi)+\frac{\xi}{2}\hat{f}(\xi)=0$$

$$\displaystyle\hat{f}(\xi)=Ce^{-\xi^2/4}=\Lambda(\xi)$$

Where it is known that $\Lambda(0)=\sqrt{\pi}$

$$\therefore\hat{f}(\xi)=\sqrt{\pi}e^{-\xi^2/4}$$

So

$$\Lambda(\lambda)=\sqrt{\pi}e^{-\lambda^2/4}$$

Community
  • 1
mallan
  • 2,009