Uniform convergence of $\int_0^{+\infty}\frac{1-\cos{\alpha x}}{x}e^{-\lambda x}dx$

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!

• Uniform convergence for $\alpha\to 0$ or $\alpha\to \infty$? – Mostafa Ayaz Jan 9 '18 at 12:51
• 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. – A.Γ. Jan 27 '18 at 12:36

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 $B=0$.
$$\int_0^{+\infty}\frac{1-\cos{\alpha x}}{x}e^{-\lambda x}dx= \int_0^{+\infty}\frac{1-\cos{x}}{x}e^{-\frac{\lambda}{\alpha} x}dx =\frac{\alpha}{\lambda}\int_0^{+\infty}\frac{\sin{x}}{x}e^{-\frac{\lambda}{\alpha} x}dx.$$