Question

Definite double integral of even function

Answer 1

Hello..........

When calculating Fourier series, you often consider integrals of the form
I=∫a−af(x)dx
I=∫−aaf(x)dx
If ff is odd or even, then sometimes you can make this simpler. We can rewrite that integral in the following way:
I=∫a−af(x)dx=∫0−af(x)dx+∫a0f(x)dx=∫a0f(−x)dx+∫a0f(x)dx
I=∫−aaf(x)dx=∫−a0f(x)dx+∫0af(x)dx=∫0af(−x)dx+∫0af(x)dx
For an even function, we have f(−x)=f(x)f(−x)=f(x), whence
I=2∫a0f(x)dx
I=2∫0af(x)dx
For an odd function, we have f(−x)=−f(x)f(−x)=−f(x), whence
I=∫a0(−f(x)+f(x))dx=0
I=∫0a(−f(x)+f(x))dx=0
That’s what it means to simplify the integration: the integral of an odd or even function over the interval [−a,a][−a,a] can be put into a nicer form

Hope this would help you
@Rohit

Answer 2

Even Function

A univariate function  is said to be even provided that . Geometrically, such functions are symmetric about the -axis. Examples of even functions include 1 (or, in general, any constant function)

An even function times an odd function is odd, while the sum or difference of two nonzero functions is even if and only if each summand function is even. The product or quotient of two even functions is again even.

If a univariate even function is differentiable, then its derivative is an odd function; what's more, if an even function is integrable, then its integral over a symmetric interval, is precisely the same as twice the integral over the interval . Similarly, if an odd function is differentiable, then its derivative is an even function while the integral of such a function over a symmetric interval  is identically zero.

Ostensibly, one can define a similar notion for multivariate functions by saying that such a function is even if and only if

Even so, such functions are unpredictable and very well may lose many of the desirable geometric properties possessed by univariate functions. For example, both and  satisfy this identity while the constant slices and of  an are odd and even, respectively. Differentiability and integrability properties are similarly unclear.

The Maclaurin series of an even function contains only even powers.

hope it helps u buddy ✌