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r(x) = x if x \geq 0 and r(x) = 0 if x<0
I have to show that:
1-\[ \int_{- \infty}^{+ \infty} r(x) \varphi ''(x) dx = \varphi(0) \]
And 2- that the second derivative of r is the Dirac delta.
And I managed to do this by integrating by parts. Howver, I don't get why I can't just write:
\[ \int_{- \infty}^{+ \infty} r''(x) \varphi (x) dx = \varphi(0) \]
Wouldn't that be correct considering distributions (I actually used this to show the second point)? I guess my question is, why can't I write the second derivatives of the ramp function (the derivative of the Heaviside function) simply as
r(x) = 0 if x \geq 0 and r(x) = 0 if x<0
i.e. 0
Which would make the integral = 0
Does it only have a second derivative in the distribution sense? Why?
EDIT: I don't get why my message is being displayed like this...
I have to show that:
1-\[ \int_{- \infty}^{+ \infty} r(x) \varphi ''(x) dx = \varphi(0) \]
And 2- that the second derivative of r is the Dirac delta.
And I managed to do this by integrating by parts. Howver, I don't get why I can't just write:
\[ \int_{- \infty}^{+ \infty} r''(x) \varphi (x) dx = \varphi(0) \]
Wouldn't that be correct considering distributions (I actually used this to show the second point)? I guess my question is, why can't I write the second derivatives of the ramp function (the derivative of the Heaviside function) simply as
r(x) = 0 if x \geq 0 and r(x) = 0 if x<0
i.e. 0
Which would make the integral = 0
Does it only have a second derivative in the distribution sense? Why?
EDIT: I don't get why my message is being displayed like this...
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