2
$\begingroup$

Please help me to solve of the following problem:

Let $\alpha, \lambda>0$. Prove uniform convergence of improper integral with respect to $\alpha$: $$\int_0^{+\infty}\frac{1-\cos{\alpha x}}{x}e^{-\lambda x}dx$$

My attempt:

I think I should use Dirichlet's test. Is it correct? My notes contain the following variant:

Let $f(x,\alpha), g(x, \alpha)$ are defined on $[0,+\infty)\times(0, +\infty)$. Then improper integral $\int_0^{+\infty}f(x,\alpha) g(x, \alpha)\ dx$ converges uniformly on $(0,+\infty)$ if the following conditions hold:

  1. $f(x,\alpha), g(x, \alpha)$ are continuous on $[0,+\infty)\times(0, +\infty)$.

  2. $|\int_0^{B}f(x,\alpha)\ dx|< C$ for some $C>0$ and for all $B>0$ and $\alpha \in (0,+\infty)$.

  3. $g(x, \alpha)$ is monotone with respect to $x$ for any fixed $\alpha$ and $g(x,\alpha)\to0$ uniformly when $x \to +\infty$ and $\alpha \in (0,+\infty)$.

Ok, next thing is to get $f$ and $g$. I think the should take $f(x,\alpha)=\frac{1-\cos{\alpha x}}{x}$ and $g(x,\alpha)=e^{-\lambda x}$. Is it correct?

Ok, now I can claim that condition $3$ is satisfied since $g(x, \alpha)$ is monotone with respect to $x$ and does not depend on $\alpha$ and $g(x, \alpha) \to 0$ as $x \to +\infty$. I also claim that condition $2$ holds as well. Is it correct?

I hope the things I stated above are correct, if I made mistakes please let me know.

Now let us move the harder part.

  1. $g(x, \alpha)$ is clearly continuous. But what about $f(x, \alpha)$? We have a problem at $x=0$. But $f(x, \alpha)=\frac{\sin^2(\frac{\alpha x}{2})}{x},$ so $f(x, \alpha)$ has nice limit $0$ as $x \to 0$. Taking this into account, can we still use Dirichlet's test and claim that condition 1 is safisfied? How can we justify that? Do not spend your time is the proof is long, just please prove a reference.

  2. What about condition $2$? I did not make any progress with it.

If my approach with Dirichlet's test is wrong, please let me know and suggest working approach.

Thanks a lot for your help!

$\endgroup$
  • $\begingroup$ Uniform convergence for $\alpha\to 0$ or $\alpha\to \infty$? $\endgroup$ – Mostafa Ayaz Jan 9 '18 at 12:51
  • $\begingroup$ I like the way how you ask questions - with details, thoughts and attempts. They make me willing to help even when I am busy with something else. $\endgroup$ – A.Γ. Jan 27 '18 at 12:36
0
$\begingroup$

Condition 2 is not fulfilled because, loosely speaking, on average $\sin ^2(\frac{\alpha x}2)$ equals $\frac 12$ and the integral of $\frac 1{2x}$ diverges for $x\rightarrow\infty$.

It might be easier to fall back on the definition of uniform convergence and find an upper bound for $\int_B^\infty \frac{1-\cos \alpha x}xe^{-\lambda x}dx$ that does not depend on $\alpha$ and converges to $0$ for $B\rightarrow\infty$. There is no need to start at $B=0$.

$\endgroup$
0
$\begingroup$

$$\int_0^{+\infty}\frac{1-\cos{\alpha x}}{x}e^{-\lambda x}dx= \int_0^{+\infty}\frac{1-\cos{x}}{x}e^{-\frac{\lambda}{\alpha} x}dx =\frac{\alpha}{\lambda}\int_0^{+\infty}\frac{\sin{x}}{x}e^{-\frac{\lambda}{\alpha} x}dx. $$

Now can erasily apply your theorem

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.