# Math Help - Integrating Sin(x)/x

1. ## Integrating Sin(x)/x

Hey I've been having trouble integrating:

$\int_0^1 \frac{\sin x}{x} dx$

Here's what I have tried so far:

$I = \int_0^1 \underbrace{\frac{1}{x}}_{\tfrac{d}{dx} \ln x} \sin x dx = \underbrace{\left. \ln x \cdot \sin x\right|_0^1}_0 - \int_0^1 \ln x \cdot \cos x dx$

$u = \ln x \rightarrow du = \frac{dx}{x} \ \ \ [Note: x = e^u]$

But here's where I encounter a problem...I can't break it down properly after I do that substitution ... I just keep getting 0 = 0 or something like that. Am I on the correct path though?

2. You can't integrate $\frac{\sin x}x$ but you can use power series for $\sin x$ to approximate a value for such integral.

3. That could explain why I couldn't integrate it.

Could I use a Taylor Series for this?

And how would doing a series for sin x help get sinx/x?

4. For all $x,$

$\sin x = \sum\limits_{k = 0}^\infty {\frac{{( - 1)^k x^{2k + 1} }}
{{(2k + 1)!}}}.$
From there you can get an expression for the integrand.

Substitute that into the integral, interchange the sum by the integral and the rest follows.

5. Originally Posted by Krizalid
For all $x,$

$\sin x = \sum\limits_{k = 0}^\infty {\frac{{( - 1)^k x^{2k + 1} }}
{{(2k + 1)!}}}.$
From there you can get an expression for the integrand.

Substitute that into the integral, interchange the sum by the integral and the rest follows.
Thanks...

Weirdly this is the exact same thing we were doing today in class.

I'm not sure how to go about getting an expression from that for the integrand though. I obviously have to do something to it to get rid of the k???

6. Originally Posted by TrevorP
Thanks...

Weirdly this is the exact same thing we were doing today in class.

I'm not sure how to go about getting an expression from that for the integrand though. I obviously have to do something to it to get rid of the k???
$\sin x = \sum\limits_{k = 0}^\infty {\frac{{( - 1)^k x^{2k + 1} }}
{{(2k + 1)!}}}.$

$\Rightarrow \frac {\sin x}x = \sum\limits_{k = 0}^\infty {\frac{{( - 1)^k x^{2k} }}
{{(2k + 1)!}}} = 1 - \frac {x^2}{3!} + \frac {x^4}{5!} - \frac {x^6}{7!} + ...$

integrate (the thing in the middle) term by term

7. Originally Posted by TrevorP
Hey I've been having trouble integrating:

$\int_0^1 \frac{\sin x}{x} dx$

Here's what I have tried so far:

$I = \int_0^1 \underbrace{\frac{1}{x}}_{\tfrac{d}{dx} \ln x} \sin x dx = \underbrace{\left. \ln x \cdot \sin x\right|_0^1}_0 - \int_0^1 \ln x \cdot \cos x dx$

$u = \ln x \rightarrow du = \frac{dx}{x} \ \ \ [Note: x = e^u]$

But here's where I encounter a problem...I can't break it down properly after I do that substitution ... I just keep getting 0 = 0 or something like that. Am I on the correct path though?
The function you want to integrate is famous enough to have its own name. You might find the following links interesting:

sinc 1

sinc 2

8. Originally Posted by mr fantastic
The function you want to integrate is famous enough to have its own name. You might find the following links interesting:

sinc 1

sinc 2
Just because the function has no primitive it does not need the integral has no closed form. If it limits were from 0 to oo then it does have closed form.

9. Originally Posted by ThePerfectHacker
Just because the function has no primitive it does not need the integral has no closed form. If it limits were from 0 to oo then it does have closed form.
Actually the actual question does have 0 to infinity. I just was going to break it up into two parts. (0 - 1, 1 - infinity)

But are you saying that there is a way to do this with 0 - infinity?

10. Originally Posted by TrevorP

But are you saying that there is a way to do this with 0 - infinity?
If it is from 0 to oo there are many ways to do it. I post one soon.

11. Originally Posted by TrevorP
Actually the actual question does have 0 to infinity. I just was going to break it up into two parts. (0 - 1, 1 - infinity)

But are you saying that there is a way to do this with 0 - infinity?
Yes, and the answer is $\frac{\pi}{2}$. As TPH said, there are several ways this can be shown. If I have time before TPH does, I'll post one.

12. Originally Posted by TrevorP
Actually the actual question does have 0 to infinity.
Aha! This makes more sense!

As a first solution, we can easily tackle this with double integration.

Consider $\int_{ - \infty }^{ + \infty } {\frac{{\sin x}}
{x}\,dx} .$
Besides $\frac{1}
{x} = \int_0^\infty {e^{ - ux} \,du} .$
We're going to create a double integral, and of course, the only reason that I'd ever create a double integral that wasn't there in first place is so I could reverse integration order. (This is justified by Tonelli's Theorem.)

Since the function we're tryin' to integrate is even, after insert the parameter and reverse integration order we have:

$\int_{ - \infty }^{ + \infty } {\frac{{\sin x}}
{x}\,dx} = 2\int_0^\infty {\int_0^\infty {e^{ - ux} \sin x\,dx} \,du} .$

Now let's make this easier with complex numbers:

$\int_0^\infty {e^{ - ux} \sin x\,dx} = \text{Im} \int_0^\infty {e^{ - (u - i)x} \,dx} = \text{Im}\, \frac{1}
{{u - i}} = \frac{1}
{{u^2 + 1}}.$

Hence $\int_{ - \infty }^{ + \infty } {\frac{{\sin x}}
{x}\,dx} = \int_0^\infty {\frac{2}
{{u^2 + 1}}\,du} ,$
and the rest follows.

13. Wow that's cool.

It's gonna take me a minute to look at each step and figure out what you did. (especially since we haven't done any Imaginary Number rules yet.)

But I appreciate it...this is the kind of math I like!

14. Actually this is the best way to kill that integral.

As you're expected to know, integration by parts is the straightforward way to do it, but it's too nasty. (At least for me.)

---

Recall that $\sin x=\text{Im}\,(e^{ix})$ & $\cos x=\text{Re}\,(e^{ix}).$ (This is actually Euler's Formula.)

15. Yeah we definitely haven't done Euler's yet ... but I guess that explains why this in on the wall in my engineering lounge:

$e^{wi} = \cos(w) + i\sin(w)$

Yeah my question gave us the hint of Integration by parts...but that didn't help me at all.

I like this answer much more. I might not actually hand it in just because it's obviously something we wouldn't know yet. But it's still interesting to know.

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