Antiderivative of an even function

[filter54321]

Is the antiderivative of an even function even? Specifically, given an even function g(x)=g(-x) will the following function be even? Odd?

For constant a>0, r a dummy variable

h(x,t) = Definite Integral [ g(r) dr, FROM x-at TO x+at]

Asking whether or not h(x,t)=h(-x,t)


This is the last bit to a larger PDE problem and it's not coming to me...

 

[filter54321]

Ignore the first part. Sine/cosine is a simple counterexample. Still wondering about the second part.

Thanks

 

[mathman]

If the function is even, yes. [x-at,x+at] and [-x-at,-x+at] cover mirror images of the real line.

 

[AlephZero]

$$h(x,t) = \int_{x-at}^{x+at}g(r)\,dr$$

and $g$ is even

$$h(x,-t) = \int_{x+at}^{x-at}g(r)\,dr = -h(x,t)$$

$$h(-x,-t) = \int_{-x+at}^{-x-at}g(r)\,dr$$

Let $s = -r$, $ds = -dr$

$$h(-x,-t) = -\int_{x-at}^{x+at}g(-s)\,ds = -h(x,t)$$

So

$$h(-x,t) = -h(-x,-t) = h(x,t)$$