ò2x1/2 sin(x) dx
By Anonymous on Friday, December 29,
2000 - 10:54 pm :
How about this?
ò2x1/2 sin(x) dx?
By Brad Rodgers (P1930) on Saturday,
December 30, 2000 - 02:59 am :
As far as I can tell, evaluating the integral of
2x1/2 sin(x) with respect to dx is going to involve
the fresnel cosine integral. In fact, solving this would actually
solve the fresnel cosine integral, which I can't find a solution
for anywhere, so I don't think that there is a solution. Here's
what I did to show this:
y=2 cos(x)(x/(2p))1/2
Then,
dy/dx=-sin(x)2(x/2p)1/2+cos(x)/(2px)1/2 Therefore,
ò2x1/2 sin(x) dx=(2p)1/2[-cos(x)(2x/p)1/2+òcos(x)/ (2p×x)1/2 dx]
If we have a solution to
òcos(x)/(2px)1/2 dx
then we have a solution to the fresnel integral:
m=òcos(pt2/2) dt
as if we let
x=pt2/2
then
dx=pt dt =(2px)1/2 dt
and we obtain,
dm=cos(x)/(2px)1/2 dx
so
òcos(x)/(2px)1/2 dx=òcos(pt2/2) dt
Which is the term we need to find. So, I don't think that the result can be
simplified any further. In terms of the fresnel integral, the answer though is
-2x1/2 cos(x)+(2p)1/2×C((2x/p)1/2)
where C is the fresnel cosine integral.
This all of course can be simplified if there is an easy solution
to the fresnel integral.
Brad
By Anonymous on Saturday, December 30,
2000 - 02:42 pm :
Thanks Brad!
I've never heard of a Fresnel Integral before? Can you explain a
little detail of what it means? And when it is to be used. Maybe
if you've got time, some small examples on what a Fresnel
Integral is also.
By Brad Rodgers (P1930) on Saturday,
December 30, 2000 - 05:10 pm :
I actually just found out about it last night while looking
for a way to integrate your expression. Anyways, here's what I
was able to find:
C(x) [the fresnel cosine integral] is defined as
C(x)=ò0x cos(pt2/2) dt
and
S(x) [the fresnel sine integral] is defined as
S(x)=ò0x sin(pt2/2) dt
The application my book gives here is if the error function is
erf(t)=2/(p)1/2ò0t e-y2 dy
then
(1-i)-1erf(t)=C(t)-i S(t)
where i is imaginary 1. I'm afraid I don't know a proof.
Directly above this is, "Fresnel Diffraction"; so I think it may
have some applications to diffraction, but surprisingly, it says
nothing about this in the book.
The only application that I can find other than the error
function and possibly diffraction is soving integrals like the
one you have. Since you can numerically find aproximate areas
under a function, the answer to this problem is very much like a
standard function. I'll try to find out more about this, sorry
that I couldn't be of more help. Perhaps someone else knows
more...
Brad
By Anonymous on Saturday, December 30,
2000 - 05:18 pm :
Thank you Brad!
By Sean Hartnoll (Sah40) on Saturday,
December 30, 2000 - 06:53 pm :
Brad - the proof you mention is I suspect just using
ei x=cos x+i sin x, maybe followed by a change of variables.
There are various series expansions for the error function. There are also
so-called asymptotic expansions which give a better approximation for large
x.
One famous (very useful) result that is not too difficult to prove, although
not trivial (perhaps you could try) is that
Sean
By Sean Hartnoll (Sah40) on Saturday,
December 30, 2000 - 06:56 pm :
I should have added that the reason it is famous is that
it is another mysterious link between the numbers e and p and it is
useful because the curve f(x)=e-x2 is called the Gaussian or Bell or
Normal curve and is used all over the place in statistics and also quantum
mechanics and statistical physics.
Sean
By Brad Rodgers (P1930) on Tuesday,
January 2, 2001 - 07:10 pm :
How do you come up with that result for the bell curve? I've
tried a few different ways:
1. Write in terms of the error function. If the error function is
equal to 1 at infinity, then the result is true. But this just
leads to a harder problem unless there is an easy way to evaluate
the error function.
2. Solve the indefinite integral straight forward. This appears
to be impossible. Or at least very hard.
3. I had the idea of expanding it into a series, but the only way
I know to expand a series is way of the Maclaurin series I
get:
òe-y2dy=(y)-(y3/3)+(y5/5×2!)-(y7/7×3!)+(y9/9!×4!)-...
I have no idea how to evaluate this, or even it is able to be
evaluated. You mentioned something about asyptotic expansions
being accurate for larger values of y. What are these, and would
they work better to solve this sort of equation?
4. My last idea was to put -y2 =ix
This gives the integral we need to evaluate as
(i1/2/2)ò[cos(x)+i×sin(x)]/x1/2
This of course involves the Fresnel Integral, which doesn't seem
to be an easier integral to evaluate either.
Can you give a little hint as to what I need to do to sove this
integral?
Thanks,
Brad
By Sean Hartnoll (Sah40) on Tuesday,
January 2, 2001 - 11:28 pm :
Brad - It's probably best if I just give you the answer,
it's probably a bit tricky to get on your own, and you may not have seen all
the steps before
We want to evaluate I = ò-¥¥ e-x2 dx.
Consider
I2 = ò-¥¥ò-¥¥e-x2-y2 dx dy
This is called a multiple integral. To start with you can think of it as
meaning we integrate over x and then over y. But a better way to think of
it is that we are integrating the function over the whole x-y PLANE and that
dx dy is an area element (wheras before dx was a length element). The
trick now is to rewrite this expression in polar coordinates, defined by
x = r cosq y = r sinq
(which imply r2 = x2 + y2)
The area element in these coordinates is
r dqdr
(this can be motivated by noting that if you integrate r from 0 to a and
q from 0 to t then you get the area of a circular segment of radius
a and angle t).
So the integral is now
I2 = ò0¥ò02p e-r2 r dqdr
Now because there is an extra r we can do the integral! Perhaps you would
like to do it and see that we get p, which then gives the result by taking
square roots.
Let me know if any of these steps don't make sense,
Sean
By Sean Hartnoll (Sah40) on Tuesday,
January 2, 2001 - 11:31 pm :
Asymptotic expansions are expansions of
functions in terms of other functions (such as powers) such that
the difference between the function and its expansion goes to
zero as the variable x goes to infinity.
Sean
By Michael Doré (Md285) on Tuesday, January
2, 2001 - 11:37 pm :
I thought that asymptotic meant that the
ratio -> 1 rather than the difference -> 0. Of course these
are the same in this case (I think) as the function isn't tending
to 0 or infinity...
By Brad Rodgers (P1930) on Wednesday,
January 3, 2001 - 03:41 pm :
I think that I understand, although a few details are a bit foggy. The main
thing I don't understand is why r dqdr is the area element. In other
words, why is dy dx=r dr dq? Other than this it seems to make sense to
me.
Thanks,
Brad
By Sean Hartnoll (Sah40) on Thursday,
January 4, 2001 - 05:09 pm :
Okay, first the point I made in the previous post is
essentially why you want it to be true, if you take r dr dq to be the
area element then the area of a segment with radius a and angle q¢
is
Area = ò0a ò0q¢r dr dq = q¢×a2/2
which is what it should be (note that we get the area for a circle if we put
q¢ = 2p.
To see the infinitessimal case directly (draw a picture)
1) go to a point (r,q) 2) increase r slightly, by dr. 3) increase q slightly, by dq. By the formula for the length of
a circular curve, we know that this gives a length r dq.
What we get is not quite a rectangle, but as we let dr and dq get
arbitrarily small, then it is nearly a rectangle and it has area
r dqdr.
Sean