Showing Dirichlet Integral exists The problem I'm trying to do is the following, Spivak Chapter 19 Problem 43.
Problem:
a) Use integration by parts to show that
$$\int_a^b \frac{\sin x}{x}d{x}=\frac{\cos a}{a}-\frac{\cos b}{b} -\int_a^b\frac{\cos x}{x^2}d{x}$$
and conclude that $\int_0^\infty \frac{\sin x}{x}$ exists. (Use the left side to investigate the limit as $a\to 0$ and the right side for the limit as $b \to \infty$.)
c) Prove that 
$$\lim_{\lambda \to \infty}\int_0^\pi \sin{(\lambda+\frac{1}{2})t}\bigg[\frac{2}{t}-\frac{1}{\sin \frac{t}{2}}\bigg]=0$$
What I tried:
For item a, the integration by parts is easy. For the rest, I think that just pointing out that the integrand behaves nicely at $x=0$ is sufficient as elsewhere it's a continuous function (right?). But for the second thing it asks regarding the limit at infinity, I wasn't able to come up with a good explanation.
For item c, I know that the limit of the brackets as x goes to zero is zero, and I know that this somehow is to be used, but I don't see how. It gives a hint: to use the "Reimann-Lebesgue Lemma".
Please keep the answers at an appropriate level and from fundamentals so that there is no need to use fancy theorems that solve in few lines, but just take the search for the answer elsewhere.
*I found many solutions to the Dirichlet problem, but I believe none of them actually took/explained these steps.
 A: For the first part you know that the function $\cos{x}\in[-1,1]$ and therefore is bounded.
Therefore the following limits hold:
$$\lim_{x\to \infty }\frac{\cos{x}}{x}=0$$
$$\lim_{x\to \infty }\int_a^x{\frac{\cos y}{y^2}}dy\leq \lim_{x\to \infty }\int_a^x{\frac{dy}{y^2}}$$
The last integral converges for $a\neq 0$ (check it for $a<0$). Since you've proved that the integral exists for $a=0$ you have finally proved that the integral exists.
The part c can be put as:
$$\lim_{\lambda\to \infty}\mathcal{I}\left\{\int_{0}^{\pi}{\exp{[i\omega(\lambda)t]}f(t)\,dt}\right\}$$
I am not sure about how to go further.
A: For (a), notice that $\lim_{x\to 0}\frac{\sin(x)}{x}=0$ implies that $\frac{\sin(x)}{x}$ is continuous and bounded in $(0,a]$ and therefore it is integrable in $[0,a]$. 
In $[a,+\infty)$, with $a>0$,
$$\int_a^{+\infty} \frac{\sin x}{x}d{x}=\frac{\cos a}{a}-\lim_{b\to +\infty}\frac{\cos b}{b} -\int_a^{+\infty}\frac{\cos x}{x^2}d{x}.$$
which is finite because $\lim_{b\to +\infty}\frac{\cos b}{b}=0$ and 
$$\frac{|\cos x|}{x^2}\leq \frac{1}{x^2}\implies \int_a^{+\infty}\frac{|\cos x|}{x^2}d{x}\leq  \int_a^{+\infty}\frac{1}{x^2}d{x}= \frac{1}{a}<+\infty$$
(recall that absolutely integrable functions are also integrable).
Hence we may conclude that $\frac{\sin(x)}{x}$ is integrable in $[0,a]\cup [a,+\infty)=[0,+\infty)$.
As regards (c), in order to apply Riemann-Lebesgue Lemma, we have to verify that
$$f(t):=\frac{2}{t}-\frac{1}{\sin \frac{t}{2}}$$
is integrable in $[0,\pi]$ which is true because 
$$\lim_{t\to 0^+}f(t)=\lim_{t\to 0^+}\left( \frac{2}{t}-\frac{1}{ \frac{t}{2}+o(t^2)}\right)=\lim_{t\to 0^+}\frac{2}{t}\left( 1-\frac{1}{ 1+o(t)}\right)0=\lim_{t\to 0^+}\frac{2o(t)}{t(1+o(t))}=0$$
and therefore $f$ is continuous and bounded in $(0,\pi]$.
