Evaluating $\lim_{b\to\infty} \int_0^b \frac{\sin x}{x}\, dx= \frac{\pi}{2}$

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Solving the integral $\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}$?

Using the identity $$\lim_{a\to\infty} \int_0^a e^{-xt}\, dt = \frac{1}{x}, x\gt 0,$$ can I get a hint to show that $$\lim_{b\to\infty} \int_0^b \frac{\sin x}{x} \,dx= \frac{\pi}{2}.$$

score 3 · Accepted Answer · edited Mar 22 '12 at 03:38

3

Hint: $$\begin{align} \lim_{b\to \infty}\int_{0}^{b}\frac{\sin x}{x}dx &= \lim_{a,b\to \infty}\int_{0}^{b}\int_{0}^{a}e^{-xt}dt\sin x dx\\& = \lim_{a,b\to \infty}\int_{0}^{b}dt\int_{0}^{a}e^{-xt}\frac{e^{ix}-e^{-ix}}{2i} dx \\&=\lim_{a,b\to \infty}\int_{0}^{b}dt\int_{0}^{a}\frac{e^{-(t-i)x}-e^{-(i+t)x}}{2i} dx\end{align}$$.

edited Mar 22 '12 at 03:38

answered Mar 22 '12 at 03:16

Geralt of Rivia

1,274

please how did you get the last equality of your first line? – Yuri Mar 22 '12 at 03:29
i suppose you already know $ e^{ix}= \cos x+i \sin x $ ? – Geralt of Rivia Mar 22 '12 at 03:32
Yes I do. I think my browser is not displaying properly. – Yuri Mar 22 '12 at 03:36
Is it just me or is the integration going to be messy. Please what's the best approach here. – Yuri Mar 22 '12 at 03:51
it's not messy at all, use your identity and another basic integration: $\int \frac{dx}{1+x^2}= \arctan x $ – Geralt of Rivia Mar 22 '12 at 03:57
you are right...What was I thinking? Thank you. – Yuri Mar 22 '12 at 04:01

score 2 · Answer 2 · 2012-03-22T03:31:57.800

2

Too long for a comment

Laplace transform: $$\mathcal{L}\left[ \frac{f(x)}{x} \right] = \int_{0}^{\infty} \frac{f(x)}{x} e^{-yx}\, dx = \int_{y}^{\infty} \mathcal{L}\left[f(x)\right]\, ds $$
Identity: $$ y = 0 \implies e^{-yx} = 1$$
Laplace transform: $$ \mathcal{L}\left[\sin(x)\right] = \frac{1}{1+s^2}$$
Integration: $$ \int \frac{1}{1+s^2}\, ds = \tan^{-1}(s)$$
Trig $$ \tan^{-1}(\infty) = \frac{\pi}{2}$$

edited Mar 22 '12 at 03:31

answered Mar 22 '12 at 03:19

wait: I need to go back and confirm Hint # 1. – Mar 22 '12 at 03:22
I think I now fixed it. – Mar 22 '12 at 03:32
Use $\displaystyle \int\limits_s^\infty e^{-mt}dm =\frac{e^{-st}}{t}$ – Pedro Mar 22 '12 at 03:32

score 2 · Answer 3 · edited Jul 28 '14 at 10:25

The usual procedure is as follows:

$$\mathcal L \left\{ \frac {\sin t} {t}\right\}(s)=\int\limits_0^\infty e^{-st}\frac {\sin t} {t}dt $$

We have that for any $f(t)$ such that the transform exists

$$\mathcal L \left\{ \frac {f(t)} {t}\right\}(s)=\int\limits_0^\infty f(t)\frac {e^{-st}} {t}dt $$

But

$$\frac {e^{-st}} {t}=\int\limits_s^\infty e^{-mt}dm$$

Under appropriate conditions we can exchange the order of the integrands and put

$$\mathcal L \left\{ \frac {f(t)} {t}\right\}(s)=\int\limits_s^\infty \int\limits_0^\infty f(t) e^{-mt}dm dt $$

This means

$$\mathcal L \left\{ \frac {f(t)} {t}\right\}(s)=\int\limits_s^\infty F(t) dt $$

where $F$ is the transform of $f$. Using this with $\sin t$ gives

$$\mathcal L \left\{ \frac {\sin t} {t}\right\}(s)=\int\limits_s^\infty \frac{1}{1+t^2} dt = \frac{\pi}{2}-\tan^{-1}s $$

$$\int\limits_0^\infty e^{-st} \frac{\sin t}{t} dt=\int\limits_s^\infty \frac{1}{1+t^2} dt = \frac{\pi}{2}-\tan^{-1}s $$

Taking $s \to 0$

$$\lim\limits_{s \to 0} \int\limits_0^\infty e^{-st} \frac{\sin t}{t}dt=\frac{\pi}{2} $$

For the last step, you need to prove that

$$\lim\limits_{s \to 0} \int\limits_0^\infty e^{-st} \frac{\sin t}{t}dt=\int\limits_0^\infty \frac{\sin t}{t}dt $$

I know you can use the dominated convergence theorem (which is not in my personal stash), and maybe some other theorems, but I'm unable to prove it, though I know it is legitimate (the exponential usually makes things work.)

Evaluating $\lim_{b\to\infty} \int_0^b \frac{\sin x}{x}\, dx= \frac{\pi}{2}$

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