APM346-2018S > Web Bonus Problems

Week 13 -- BP2

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Victor Ivrii:
Prove that if $f=\ln |x|$ then
\begin{equation*}
f'(\varphi)= pv \int x^{-1}\varphi (x)\,dx ,
\end{equation*}
where $f'$ is understood in the sense of distributions and the integral is understood as a principal value integral.

Adam Gao:
$f'(x) = x^{-1}$

$f'(\varphi) = \int f'(x) \varphi(x)\,dx = \int x^{-1}\varphi (x)\,dx = pv\int x^{-1}\varphi (x)\,dx$

Concern/Confusion: Since the derivative of log absolute value of x is same as the derivative of log of x, why do we use a principal value integral instead of a normal integral? Did I somehow abuse notation?

Victor Ivrii:
Because in $\varphi(0)\ne 0$ even improper $\int x^{-1}\varphi(x)\,dx$ does not exist. Because of this all your arguments are wrong. Not wrong-wrong-wrong, but wrong.

Adam Gao:
Sorry, I am still confused. Why do we assume $\varphi(0)\ne 0$?

Would I be starting in the right direction with this instead?

$f'(\varphi) = f(\varphi ') = \int f(x) \varphi'(x)\,dx = \int \mathrm{ln}|x| \varphi'(x)\,dx$

Victor Ivrii:

--- Quote from: Adam Gao on April 04, 2018, 03:30:03 PM ---Sorry, I am still confused. Why do we assume $\varphi(0)\ne 0$?

Would I be starting in the right direction with this instead?

$f'(\varphi) = f(\varphi ') = \int f(x) \varphi'(x)\,dx = \int \mathrm{ln}|x| \varphi'(x)\,dx$

--- End quote ---
Why do you assume that $\varphi(0)= 0$?? We should try any test functions, including those with $\varphi(0)\ne 0$.

You started in the correct direction but you cannot integrate by parts in $\int_{-\infty}^\infty$. Instead you need to look what vp means

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