Explicit Integrals and LimInf/LimSup (a) Show that $f(t):=\int_0^\infty e^{-tx}\frac{sin \space x}{x}dx$ exists for $t>0$ and defines a differentiable function $f$. Calculate $f'(t)$ for $t>0$ and evaluate it explicitly.
(b) Prove that $lim_{t\rightarrow 0}\space f(t)$ exists and evaluate it explicitly.
(c)We know that $\frac{sin \space x}{x}$ is NOT Lebesgue integrable over $\mathbb{R}_+$, so $f(0)$ is NOT defined by (a). But prove that for every $N>0$,$$\lim_{t\to0^+}\space sup\space  f(t)\leq \int_0^N\frac{sin \space x}{x}dx\space + \space \frac{1}{N}$$Prove a similar formula with $lim\space inf$ as well.
(d)Evaluate $\lim_{N\to\infty}\int_0^N\frac{sin \space x}{x}dx$explicitly, based on preceding parts.
I haven't ever really dabbled with proofs and it has been a long time since I did any type of calculus (I,II, or III for that matter). Is there something I'm not getting here? 
 A: The alternative series test that you no doubt learned in Calculus is a good tool for dealing with this problem because it gives you error estimates. Recall that if $a_n \ge 0$ satisfies
$$
         a_0 \ge a_1 \ge a_2 \ge a_3 \ge a_4 \ge \cdots ,\;\;\; a_n \rightarrow 0,
$$
Then $\sum_{n=0}^{\infty}(-1)^{n}a_n$ converges, and the maximum error in the value when you truncate the sum is no larger than the first discarded term:
$$
            \left|\sum_{n=0}^{\infty}(-1)^na_n -\sum_{n=0}^{N}(-1)^n a_n\right|  \le a_{N+1}.
$$
This is useful in this case because it gives the convergence of
$$
       \int_{0}^{\infty}e^{-tx}\frac{\sin x}{x}dx = \sum_{n=0}^{\infty}\int_{n\pi}^{(n+1)\pi}e^{-tx}\frac{\sin x}{x}dx,\;\;\; t \ge 0.
$$
And it gives the following uniform error estimate
\begin{align}
      & \left|\int_{0}^{\infty}e^{-tx}\frac{\sin x}{x}dx-\int_{0}^{n\pi }e^{-tx}\frac{\sin x}{x}dx\right| \\
    & \le \int_{n\pi}^{(n+1)\pi}e^{-tx}\left|\frac{\sin x}{x}\right|dx \\
    & \le \int_{n\pi}^{(n+1)\pi}\frac{1}{x}dx \\
    & \le \frac{1}{n\pi}\pi  = \frac{1}{n}
\end{align}
The functions
$$
              f_n(t) = \int_{0}^{n\pi}e^{-tx}\frac{\sin x}{x}dx
$$
are continuous on $[0,\infty)$, and these functions converge uniformly as $n\rightarrow\infty$ to $\int_{0}^{\infty}e^{-tx}\frac{\sin x}{x}dx$, which proves that the limit function is also continuous on $[0,\infty)$. Therefore,
$$
          \lim_{t\rightarrow 0}\int_{0}^{\infty}e^{-tx}\frac{\sin x}{x}dx =\lim_{n\rightarrow\infty}\int_{0}^{n\pi}\frac{\sin x}{x}dx.
$$
That should get you started.
A: The improper integral is convergent since we have for all $x >0$
$$\left|\frac{\sin x}{x}e^{-tx}\right| \leqslant e^{-tx},$$
and for $t > 0$
$$\int_0^\infty e^{-tx} \, dx = \frac{1}{t} < \infty.$$
To prove differentiability with respect to $t$, it is sufficient to show uniform convergence of
$$\int_0^\infty \frac{\partial}{\partial t}\left(e^{-tx} \frac{\sin x}{x}\right) \, dx= -\int_0^\infty e^{-tx} \sin x \, dx.$$
This follows from the Weierstrass test. For any $c > 0$ and all $t \in [c, \infty)$ we have $|e^{-tx} \sin x| \leqslant e^{-cx}$ and $e^{-cx}$ is integrable.
Hence, for $t > 0,$
$$f'(t) = -\int_0^\infty e^{-tx} \sin x \, dx.$$
Integrate twice by parts to obtain  $f'(t)$ explicitly and solve for $f(t)$ to proceed.
In particular, you will be able to compute $\lim_{t \to 0+}f(t)$ after you prove that 
$$\lim_{t \to \infty}\int_0^{\infty}e^{-tx} \frac{\sin x }{x} \, dx = 0$$
