Getting two different values for f(0)

  • #1
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Main Question or Discussion Point

Let ##\displaystyle f(t) = \int_0^{\infty} \frac{-\sin (tx) x}{x^2+1}~ dx##. It seems clear that ##f(0) = 0##.

Now, I will do some manipulations to get a different expression: ##\displaystyle f(t) = \int_0^{\infty} \frac{-\sin (tx) x}{x^2+1}~ dx = \int_0^{\infty} \frac{-\sin (tx) x^2}{(x^2+1)x}~ dx = \int_0^{\infty} \frac{-\sin (tx) (x^2+1) + \sin (tx)}{(x^2+1)x}~ dx = \int_0^{\infty} \frac{- \sin(tx)}{x} ~ dx + \int_0^{\infty} \frac{\sin (tx)}{(x^2+1)x}~ dx = -\frac{\pi}{2} + \int_0^{\infty} \frac{\sin (tx)}{(x^2+1)x}~ dx##. So here, we see that ##f(0) = - \frac{\pi}{2}##.

What am I doing wrong? Why am I getting two different values?
 

Answers and Replies

  • #2
tnich
Homework Helper
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336
Let ##\displaystyle f(t) = \int_0^{\infty} \frac{-\sin (tx) x}{x^2+1}~ dx##. It seems clear that ##f(0) = 0##.

Now, I will do some manipulations to get a different expression: ##\displaystyle f(t) = \int_0^{\infty} \frac{-\sin (tx) x}{x^2+1}~ dx = \int_0^{\infty} \frac{-\sin (tx) x^2}{(x^2+1)x}~ dx = \int_0^{\infty} \frac{-\sin (tx) (x^2+1) + \sin (tx)}{(x^2+1)x}~ dx = \int_0^{\infty} \frac{- \sin(tx)}{x} ~ dx + \int_0^{\infty} \frac{\sin (tx)}{(x^2+1)x}~ dx = -\frac{\pi}{2} + \int_0^{\infty} \frac{\sin (tx)}{(x^2+1)x}~ dx##. So here, we see that ##f(0) = - \frac{\pi}{2}##.

What am I doing wrong? Why am I getting two different values?
You assume that ##\int_0^{\infty} \frac{- \sin(tx)}{x} ~ dx = -\frac{\pi}{2}## when ##t=0##. But ##\sin(0x) = 0##.
 
  • #3
FactChecker
Science Advisor
Gold Member
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You assume that ##\int_0^{\infty} \frac{- \sin(tx)}{x} ~ dx = -\frac{\pi}{2}## when ##t=0##. But ##\sin(0x) = 0##.
I agree, especially considering the first line of the original post. How are the two different?
 
  • #4
mathman
Science Advisor
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419
[tex]\int_0^{\infty}\frac{sin(tx)}{x}dx\ and\ \int_0^{\infty}\frac{sin(tx)}{(x^2+1)x}dx[/tex] are both discontinuous at t=0, and are both =0 at that point.
 

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