Can we use integration by parts for improper integrals?

[daudaudaudau]

What's up with this

<br /> \int_{-\infty}^\infty \sin{x}\frac{1}{x}dx=\pi<br />

Now I try integration by parts
<br /> \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<br />

Why don't I get the same result?

 

[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.

 

[daudaudaudau]

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

 

[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

 

[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.

 

[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?)

 

[HallsofIvy]

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?)

The "delta function" is not an integrable function.

 

[daudaudaudau]

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

 

[tiny-tim]

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

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

 

[daudaudaudau]

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

 

[tiny-tim]

integration by parts always works.

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

 

[daudaudaudau]

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

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

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

 

[tiny-tim]

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

 

[daudaudaudau]

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

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

<br /> \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<br />

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