# Getting two different values for f(0)

## 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?

tnich
Homework Helper
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 $\sin(0x) = 0$.

FactChecker
You assume that $\int_0^{\infty} \frac{- \sin(tx)}{x} ~ dx = -\frac{\pi}{2}$ when $t=0$. But $\sin(0x) = 0$.
$$\int_0^{\infty}\frac{sin(tx)}{x}dx\ and\ \int_0^{\infty}\frac{sin(tx)}{(x^2+1)x}dx$$ are both discontinuous at t=0, and are both =0 at that point.