Explicit Integrals and LimInf/LimSup

Question

(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?

Answer 1

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

$$\lim_{t \to \infty}\int_0^{\infty}e^{-tx} \frac{\sin x }{x} \, dx = 0$$

Answer 2

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.