Integration of g[ f(x) ]

[Dave]

Could somebody please show me a general way to integrate expressions of the form

g[ f(x) ]

Thank you very much in advance.

Dave.

[Jon and Mary Miller]

Dave wrote:

>Could somebody please show me a general way to integrate expressions of the form
>
>g[ f(x) ]
>
>

No. We would if we could, but we can't. We are old and rusty, and our
wheels are tired.

But there is a general way of integrating g[f(x)]*f ' (x). I think I
can, I think I can, I think I can.

Jon Miller

[Giuseppe Bilotta]

Dave wrote:
> Could somebody please show me a general way to integrate expressions of the form
>
> g[ f(x) ]

If you happen to get by a solution and you're not 40 years old, apply
for the Fields medal.

--
Giuseppe "Oblomov" Bilotta

Can't you see
It all makes perfect sense
Expressed in dollar and cents
Pounds shillings and pence
(Roger Waters)

[David Ziskind]

Dave <aaawe...@yahoo.com> wrote in article
<5de90f03.03090...@posting.google.com>...

> Could somebody please show me a general way to integrate expressions

> of the form g[ f(x) ] ...

This can be done by a simple substitution into the basic Change of
Variable (COV) formula.

To deal with the reduced character set, we will write: Vf, for the
inverse of f. That is, (Vf) o f = f o (Vf) = identity-function.

By COV, Int{x=a to b: h(g(x))*g'(x)} = Int{u=g(a) to g(b): h(u)} .

Now, make the substitution, h(x) = f(x)/g'((Vg)(x)) . We have:

Int{x=a to b: [f(g(x))/g'((Vg)(g(x)))]*g'(x)}
= Int{u=g(a) to g(b): f(u)/g'((Vg)(u))}

and since (Vg)(g(x)) = x;

Int{x=a to b: f(g(x))} = Int{u=g(a) to g(b): f(u)/g'((Vg)(u))}

which is the desired formula.

It is worthwhile to note that a third method of expressing COV is
obtained by making the substitution (in the last formula):
f = p o (Vg) -- thus obtaining a COV formula in which the left hand
side is simply, integral over x of p(x).

David Ziskind
zis...@ntplx.net