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]

[tex]h(x,t) = \int_{x-at}^{x+at}g(r)\,dr[/tex]

and [itex]g[/itex] is even

[tex]h(x,-t) = \int_{x+at}^{x-at}g(r)\,dr = -h(x,t)[/tex]

[tex]h(-x,-t) = \int_{-x+at}^{-x-at}g(r)\,dr[/tex]

Let [itex]s = -r[/itex], [itex]ds = -dr[/itex]

[tex]h(-x,-t) = -\int_{x-at}^{x+at}g(-s)\,ds = -h(x,t)[/tex]

So

[tex]h(-x,t) = -h(-x,-t) = h(x,t)[/tex]

 

[CellCoree]

I can confirm that the antiderivative of an even function is indeed even. This is because the antiderivative of an even function g(x) is defined as the function F(x) such that F'(x) = g(x). Since an even function is symmetric about the y-axis, its derivative will also be symmetric, and therefore the antiderivative will also be symmetric.

Now, to answer the question about the function h(x,t), we must first understand what it represents. The definite integral in the expression for h(x,t) is evaluating the area under the curve of the even function g(x) from x-at to x+at. This means that the value of h(x,t) will depend on the difference between the function values at x-at and x+at, which will be the same regardless of whether we plug in x or -x. Therefore, h(x,t) will also be an even function.

In conclusion, the function h(x,t) will be even because it is the antiderivative of an even function and the input variable x does not affect the symmetry of the function. This can also be mathematically proven by substituting -x for x in the expression for h(x,t) and showing that the resulting function is still equal to h(x,t). I hope this helps in solving the larger PDE problem you are working on.