Can we use integration by parts for improper integrals?

In summary, the conversation discussed the evaluation of the integral of sin(x)/x from -∞ to ∞, and the use of integration by parts to solve it. However, due to a singularity at x=0, the result obtained was an indeterminate form of ∞ - ∞. It was pointed out that the integral must be evaluated in a different way, such as considering the interval (0,∞) and doubling the result due to the even nature of the integrand. The conversation also touched on the use of the "delta function" and the concept of indeterminate forms.
  • #1
daudaudaudau
302
0
What's up with this

[tex]
\int_{-\infty}^\infty \sin{x}\frac{1}{x}dx=\pi
[/tex]

Now I try integration by parts
[tex]
\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
[/tex]

Why don't I get the same result?
 
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  • #2
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
Yes that works, but what did I do wrong in my initial calculation? I just used integration by parts.
 
  • #4
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
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
um the integrand in

[tex]\int_{-\epsilon}^{\epsilon}\delta(x)dx[/tex]

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
ice109 said:
um the integrand in

[tex]\int_{-\epsilon}^{\epsilon}\delta(x)dx[/tex]

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.
 
  • #8
So what's wrong with the way I use integration by parts? :-)
 
  • #9
daudaudaudau said:
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-) :wink:
 
  • #10
So is there some general rule for when integration by parts works and when it fails?
 
  • #11
integration by parts always works.

what went wrong in this case was you got ∞ - ∞, which is an indeterminate form :wink:
 
  • #12
What do you mean? I got
[tex]
\int_{-\infty}^\infty \sin{x}\frac{1}{x}dx=\pi = -\int_{-\infty}^\infty \cos{x}\frac{1}{x^2}dx = \infty

[/tex]

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

[/tex]

but I don't see why, because I just used the formula for integration by parts.
 
  • #13
No, you also got [-cosx/x]-∞, = [-cosx/x]-∞0 + [-cosx/x]0, = ∞ + ∞.
 
  • #14
tiny-tim said:
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

[tex]
\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
[/tex]

where [itex] c[/itex] is any real constant. This clearly diverges for any [itex]c[/itex] ...
 

FAQ: Can we use integration by parts for improper integrals?

What is integration by parts?

Integration by parts is a mathematical technique used to find the integral of a product of two functions. It is based on the product rule of differentiation and is often used to simplify complex integrals.

When should I use integration by parts?

Integration by parts is typically used when the integral of a product of two functions cannot be easily evaluated using other integration techniques, such as substitution or trigonometric substitution.

How do I know which function to choose for u and v?

When using integration by parts, it is important to choose u and v in a way that simplifies the integral. Typically, u should be chosen as the more complicated function and v should be the derivative of u.

Are there any special cases for integration by parts?

Yes, there are two special cases for integration by parts: the tabular method and the reduction formula. The tabular method is useful for integrating products of trigonometric functions, while the reduction formula is used for recursive integrals.

Can integration by parts be used for definite integrals?

Yes, integration by parts can be used for both indefinite and definite integrals. However, when using it for definite integrals, the boundaries of integration must be adjusted accordingly.

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