# Integration by parts

1. Dec 25, 2009

### daudaudaudau

What's up with this

$$\int_{-\infty}^\infty \sin{x}\frac{1}{x}dx=\pi$$

Now I try integration by parts
$$\int_{-\infty}^\infty \sin{x}\frac{1}{x}dx=[-\cos{x}\frac{1}{x}]_{-\infty}^\infty-\int_{-\infty}^\infty \cos{x}\frac{1}{x^2}dx = -\int_{-\infty}^\infty \cos{x}\frac{1}{x^2}dx = \infty$$

Why don't I get the same result?

2. Dec 25, 2009

### mathman

There is a singularity at x=0 which cannot be ignored. To get a clearer picture, carry out the original integral (0,∞) and double it (the integrand is an even function). The integration by parts will now have two ∞'s.

3. Dec 25, 2009

### daudaudaudau

Yes that works, but what did I do wrong in my initial calculation? I just used integration by parts.

4. Dec 25, 2009

### ice109

so how did you do the first and last integrals? how do you know the first integral equals pi and how did you integrate cos/x^2

5. Dec 25, 2009

### daudaudaudau

Here's a bunch of ways for evaluating int sin(x)/x: http://www.mathlinks.ro/viewtopic.php?t=197640

The last integral is obviously divergent because the integrand diverges as 1/x^2 as x goes to zero. If you're trying to give me a hint, then I think I missed it.

6. Dec 25, 2009

### ice109

um the integrand in

$$\int_{-\epsilon}^{\epsilon}\delta(x)dx$$

diverges as x goes to zero as well the integral is 1.

now someone might call b.s. on me because really that's a definition but still my point is you can't reason just cause there's a divergence at some point that the integral around that point diverges (or can you?)

7. Dec 26, 2009

### HallsofIvy

Staff Emeritus
The "delta function" is not an integrable function.

8. Dec 26, 2009

### daudaudaudau

So what's wrong with the way I use integration by parts? :-)

9. Dec 26, 2009

### tiny-tim

Because cosx/x is discontinuous at x = 0 (it's ∞ at 0+, and -∞ at 0-)

10. Dec 26, 2009

### daudaudaudau

So is there some general rule for when integration by parts works and when it fails?

11. Dec 26, 2009

### tiny-tim

integration by parts always works.

what went wrong in this case was you got ∞ - ∞, which is an indeterminate form

12. Dec 26, 2009

### daudaudaudau

What do you mean? I got
$$\int_{-\infty}^\infty \sin{x}\frac{1}{x}dx=\pi = -\int_{-\infty}^\infty \cos{x}\frac{1}{x^2}dx = \infty$$

Somehow it must be wrong to write
$$\int_{-\infty}^\infty \sin{x}\frac{1}{x}dx= -\int_{-\infty}^\infty \cos{x}\frac{1}{x^2}dx$$

but I don't see why, because I just used the formula for integration by parts.

13. Dec 26, 2009

### tiny-tim

No, you also got [-cosx/x]-∞, = [-cosx/x]-∞0 + [-cosx/x]0, = ∞ + ∞.

14. Dec 27, 2009

### daudaudaudau

I guess you are right. I never thought about that. What I have is

$$\left[\frac{\cos{x}}{x}\right]_{-\infty}^\infty=\int_{-\infty}^\infty\frac{d}{dx}\frac{\cos{x}}{x}dx = \lim_{a\rightarrow\infty}\int_{c}^a\frac{d}{dx}\frac{\cos{x}}{x}dx+\lim_{b\rightarrow-\infty}\int_{b}^c\frac{d}{dx}\frac{\cos{x}}{x}dx$$

where $c$ is any real constant. This clearly diverges for any $c$ ...