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Prove that: $\frac{d}{dt} \int_{-\infty}^{\infty} e^{-x^2} \cos(2tx) dx=\int_{-\infty}^{\infty} -2x e^{-x^2} \sin(2tx) dx$

This is my proof:

$\forall \ t \in \mathbb{R}$ (the improper integral coverge absolutely $\forall \ t \in \mathbb{R}$) I consider: $g(t)=\int_{-\infty}^{\infty} e^{-x^2} \cos(2tx) dx$.

Let $h \ne 0$

$\left| \frac{g(t+h)-g(t)}{h}-\int_{-\infty}^{\infty} -2x e^{-x^2} \sin(2tx) dx \right|\le\int_{-\infty}^{\infty} \left|\frac{\cos(2(t+h)x)-\cos(2tx)}{h}-(-2x)\sin(2tx)\right| e^{-x^2}dx$

For the main value theorem and since $\int_{-\infty}^{\infty} e^{-x^2} dx=\sqrt{\pi}$

$\int_{-\infty}^{\infty} \left|\frac{\cos(2(t+h)x)-\cos(2tx)}{h}-(-2x)\sin(2tx)\right| e^{-x^2}dx=\left|\frac{\cos(2(t+h)\bar{x})-\cos(2t\bar{x})}{h}-(-2\bar{x})\sin(2t\bar{x})\right| \sqrt{\pi}$

$\cos(2tx)$ is derivable in $\bar{x}$ then fixed a $\epsilon>0 \ $ if $\ 0<|h|<\delta$:

$\left|\frac{\cos(2(t+h)\bar{x})-\cos(2t\bar{x})}{h}-(-2\bar{x})\sin(2t\bar{x})\right| \sqrt{\pi}<\sqrt{\pi} \ \epsilon$

It is correct?

There are other ways?

UPDATE

probably the proof is incorrect, because when I use the mean value theorem $x$ depends also from $h$ and hence the continuity of $x(h)$ is not obvious, then I can't guarantee the derivability of $\cos(2 t x(h))$ in $x$ for $h \rightarrow 0$. Am I right?

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2 Answers 2

up vote 1 down vote accepted

There is much simpler way using the Leibniz integral rule, which allows you to swap the integral and the derivation: \begin{align} \frac{d}{dt} \int_{-\infty}^{\infty} e^{-x^2} \cos(2tx) dx &= \int_{-\infty}^{\infty} \left(\frac{\partial}{\partial t} e^{-x^2} \cos(2tx) \right) dx \\ &=\int_{-\infty}^{\infty} -2x e^{-x^2} \sin(2tx) dx \end{align}


Update: As far as I know, the Leibniz rule can be used for integral $\frac{\partial}{\partial t}\int_a^b f(x,t)dx$ with possibly infinite limits $a, b$ as well, under some additional conditions. In particular, Theorem 10.3. of this article describes it:

  • there are upper bounds $|f(x,t)|\leq A(x)$ and $|\frac{\partial}{\partial t} f (x,t)| \leq B(x)$, both being independent of $t$, such that a $\int_a^b A(x) dx$ and $\int_a^b B(x) dx$ converge.

In our case we can pick $A(x)=e^{-x^2}$ and $B(x)=2|x| e^{-x^2}$.

See also Using Leibniz Integral Rule on infinite region.

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1  
Yes, but I'm not used it because the domain is not limited. (I have seen that rule for limitate domain...) Ps is a curiosity –  Madara May 2 '13 at 14:04
    
Isn't the OP trying to demonstrate this? And, as the OP said, the Leibniz rule to which you link only shows finite limits of integration. –  Ron Gordon May 2 '13 at 15:26
    
@Madara I updated the answer, hopefully explaining it. –  Petr Pudlák May 2 '13 at 16:29
    
Thanks for the sources! But this result uses Lebesgue measure or integral? So far I have studied only Riemann integral. –  Madara May 2 '13 at 16:54
    
@Madara Real analysis isn't my domain, so I could be wrong, but I'd say it doesn't matter what integral you use as long as all the integrals required by the theorem converge. –  Petr Pudlák May 2 '13 at 17:30

You way looks good. Here's an alternate way: evaluate both integrals, and see that the derivative of one equals the other.

For example,

$$\begin{align}\int_{-\infty}^{\infty} dx \, e^{-x^2} \, \cos{2 t x} &= \Re{\left [\int_{-\infty}^{\infty} dx \, e^{-x^2} e^{i 2 t x} \right ]}\\ &= e^{-t^2}\Re{\left [\int_{-\infty}^{\infty} dx \, e^{-(x-i t)^2} \right ]}\\ &= \sqrt{\pi}\, e^{-t^2}\end{align}$$

The derivative of this with respect to $t$ is

$$\frac{d}{dt} \int_{-\infty}^{\infty} dx \, e^{-x^2} \, \cos{2 t x} = -2 \sqrt{\pi} t e^{-t^2}$$

Now try with taking the derivative inside the integral:

$$\begin{align}-2 \int_{-\infty}^{\infty} dx \,x \, e^{-x^2} \,\sin{2 t x} &= -2 \Im{\left [\int_{-\infty}^{\infty} dx\,x \, e^{-x^2} e^{i 2 t x} \right ]}\\ &= -2 e^{-t^2}\Im{\left [\int_{-\infty}^{\infty} dx\,x \, e^{-(x-i t)^2} \right ]}\\ &= -2 e^{-t^2}\Im{\left [\int_{-\infty}^{\infty} dx\,(x+i t) e^{-x^2} \right ]} \\ &= -2 t \sqrt{\pi} e^{-t^2} \end{align}$$

QED

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Ok, thanks for the answer! –  Madara May 2 '13 at 14:09
    
@Madara: I've noticed that you've asked 7 questions but accepted only 1 solution. You may want to accept solutions that you have found useful if you haven't already. It helps the M.SE community know which problems have been solved to the satisfaction of the OP. –  Ron Gordon May 2 '13 at 14:45
    
Ok, thanks for the suggestion. –  Madara May 2 '13 at 14:55

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