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

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

-

## marked as duplicate by Sivaram Ambikasaran, Jack, Did, t.b., Asaf KaragilaMar 22 '12 at 12:40

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}.

-
please how did you get the last equality of your first line? –  Yuri Mar 22 '12 at 3:29
i suppose you already know $e^{ix}= \cos x+i \sin x$ ? –  Geralt of Rivia Mar 22 '12 at 3:32
Yes I do. I think my browser is not displaying properly. –  Yuri Mar 22 '12 at 3:36
Is it just me or is the integration going to be messy. Please what's the best approach here. –  Yuri Mar 22 '12 at 3:51
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 3:57

Too long for a comment

1. Laplace transform: $$\mathcal{L}\left[ \frac{f(x)}{x} \right] = \int_{0}^{\infty} \frac{f(x)}{x} e^{-yx}\, dx = \int_{y}^{\infty} \mathcal{L}\left[f(x)\right]\, ds$$

2. Identity: $$y = 0 \implies e^{-yx} = 1$$

3. Laplace transform: $$\mathcal{L}\left[\sin(x)\right] = \frac{1}{1+s^2}$$

4. Integration: $$\int \frac{1}{1+s^2}\, ds = \tan^{-1}(s)$$

5. Trig $$\tan^{-1}(\infty) = \frac{\pi}{2}$$

-
wait: I need to go back and confirm Hint # 1. –  user2468 Mar 22 '12 at 3:22
I think I now fixed it. –  user2468 Mar 22 '12 at 3:32
Use $\displaystyle \int\limits_s^\infty e^{-mt}dm =\frac{e^{-st}}{t}$ –  Pedro Tamaroff Mar 22 '12 at 3:32

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.)

-