Re: Integral derivation ...

On Mar 25, 10:57 am, Rune Allnor <all...@xxxxxxxxxxxx> wrote:
On 25 Mar, 13:29, "dvsarw...@xxxxxxxxx" <dvsarw...@xxxxxxxxx> wrote:

In this instance, a closed-form solution is not possible
except as an infinite series ...
Finally, the OP's problem can be solved by a simple
change of variables ...

Seems you are contradicting yourself here?

Rune


OK, folks, from the top.....

Given two expressions F(x) and G(x), does F(x) equal G(x)?
For example, does x^2 + 2x + 1 equal 4x? The answer
is No as mathematicians know, but engineers who grab
their calculators and check whether equality holds, say
at x = 1, might be misled. A persistent and cautious engineer
may try x = 2 next and discover the error, but time, as many
posters in this thread have pointed out, is of the essence,
and we want to get "the answer" as soon as possible and
with as little effort as possible....

Does the integral from -infinity to b of f(x) equal the integral
from h(b) to +infinity of g(x)? Just evaluating the two integrals
numerically for a specific value of the parameter b might give
the wrong answer because equality (or maybe equality to
the numerical accuracy of the integration routine) happens to
hold for that choice of the parameter b but not for other choices
of b. In the instance asked about by the OP, the numerical
value of b is not specified, and so the question is really of the
type whether functions F(b) and G(b) are equal. Now, both integrals
can be expressed as infinite series in b. Are the two series the
same? Once again, we could evaluate the series for some
chosen value of b, but could be misled. But, **for the specific
problem asked for by the OP**, the correct answer can be
easily determined *****without finding the numerical value of
either integral**** by using the change of variable method
suggested by Tim Westcott, one of the earliest responders in
this thread.

The integral from -infinity to b of (1/s)exp[-{(x-m)^2}/{2s^2}]
equals,
by a change of variables y = (x-m)/s, the integral from -infinity to
(b-m)/s of exp[-{y^2}/2] , and by a further change of variables z = -
y,
the integral from (m-b)/s of exp[-{z^2}/2]. It is to be hoped that
the OP will be able to figure out the answer to her question from
this additional information and put this thread to rest.

Sheesh!

--Dilip Sarwate
.

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