Integral of Function with no Antiderivative
david....@gmail.com
doesn't have an antiderivative. E.g. in class we found that ∫ e^(-
x^2) dx from 0 to ∞ is 1/2 × √π using a calculator, but how would one
actually go about finding this?
Denis Feldmann
Rod
for your example.
<david....@gmail.com> wrote in message
news:1174423487.9...@o5g2000hsb.googlegroups.com...
How do you calculate the value of an integral when the function
doesn't have an antiderivative. E.g. in class we found that ? e^(-
x^2) dx from 0 to ? is 1/2 × ?? using a calculator, but how would one
Michael
This particular example is well known.
Let I = integral from 0 to infinity of e^(-x^2) dx
Then I^2 = (integral from 0 to infinity of e^(-x^2) dx) * (integral
from 0 to infinity of e^(-x^2) dx)
= (integral from 0 to infinity of e^(-x^2) dx) * (integral from 0 to
infinity of e^(-y^2) dy)
= double integral from x=0 to infinity and y=0 to infinity e^(-
(x^2+y^2)) dy dx
At this point, you convert to polar coordinates r and theta, where r
goes from 0 to infinity and theta goes from 0 to pi/2 (i.e., only in
first quadrant).
I^2 = integral theta=0 to pi/2 integral r = 0 to infinity e^(-r^2) * r
dr dtheta
= pi/2 * (1/2)[1 - 0]
= pi/4
So I = 1/2 * sqrt(pi)
Michael
Passerby
>How do you calculate the value of an integral when the function
>doesn't have an antiderivative. E.g. in class we found that ? e^(-
>x^2) dx from 0 to ? is 1/2 × ?? using a calculator, but how would one
>actually go about finding this?
Please use standard ASCII. All some of us see is a ? where you have
used a special symbol.
If all you are asking is how to do the integral
Int[0..oo] e^(-x^2) dx
then see lines (4) thru (6) here:
<http://mathworld.wolfram.com/GaussianIntegral.html>
The integral of e^(-x^2) dx does have some "named" functions
associated with it, for example the error functions erf and erfc.
For more on that, see:
<http://mathworld.wolfram.com/Erf.html>
<http://mathworld.wolfram.com/Erfc.html>
Matt Bruder
news:1174425422.2...@d57g2000hsg.googlegroups.com...
>This particular example is well known.
>Let I = integral from 0 to infinity of e^(-x^2) dx
>Then I^2 = (integral from 0 to infinity of e^(-x^2) dx) * (integral
>from 0 to infinity of e^(-x^2) dx)
>= (integral from 0 to infinity of e^(-x^2) dx) * (integral from 0 to
>infinity of e^(-y^2) dy)
>= double integral from x=0 to infinity and y=0 to infinity e^(-
>(x^2+y^2)) dy dx
>At this point, you convert to polar coordinates r and theta, where r
>goes from 0 to infinity and theta goes from 0 to pi/2 (i.e., only in
>first quadrant).
>I^2 = integral theta=0 to pi/2 integral r = 0 to infinity e^(-r^2) * r
>dr dtheta
Why? I can see that r^2=x^2+y^2, but I can't see why dx*dy = r*dr*dtheta. Do
you mind explaining this step and the new bounds?
Thanks!
> = pi/2 * (1/2)[1 - 0]
> = pi/4
>So I = 1/2 * sqrt(pi)
>Michael
--
Matt
Michael
> >(x^2+y^2)) dy dx
> >At this point, you convert to polar coordinates r and theta, where r
> >goes from 0 to infinity and theta goes from 0 to pi/2 (i.e., only in
> >first quadrant).
> >I^2 = integral theta=0 to pi/2 integral r = 0 to infinity e^(-r^2) * r
> >dr dtheta
>
> Why? I can see that r^2=x^2+y^2, but I can't see why dx*dy = r*dr*dtheta. Do
> you mind explaining this step and the new bounds?
Basically: x = r cos theta, y = r sin theta
When you convert from x,y to r,theta you have to multiply by the
determinant:
| dx/dr dy/dr |
| dx/dtheta dy/dtheta |
=
| cos theta sin theta |
| -r sin theta r cos theta |
= r cos^2 theta - (-r sin^2 theta)
= r (cos^2 theta + sin^2 theta)
= r
It took me years to understand *why* you need to multiply by the
determinant. It will probably take me several more years to figure
out how to phrase it so it's comprehensible: Basically, when you're
going from dx and dy to dr and dtheta, or in general when you convert
coordinates, there's a certain stretching or shrinking, and the
determinant corrects for that. Kind of like how, in 1D calculus, if
you integrate w.r.t. x, and I integrate with respect to u = x^2, I
have to use du = 2x * dx, i.e., I have to scale to correct for the
fact that I'm using a different variable of integration.
Actually, let me take one more stab at it. Imagine you have a vector
equation of the form a = M * b:
[dx] = [ dx/dr dy/dr ] * [dr ]
[dy] [ dx/dtheta dy/dtheta ] [dtheta]
The typography is awkward, but it should be clear: the left hand side
is a vector of 2 components where the first is dx and the second is
dy. Etc.
Now we can write M = det(M) * M', where M' rotates the coordinates but
doesn't stretch or shrink them, so it preserves lengths. So to
convert coordinates, you need to multiply by det(M), to correct for
stretching/shrinking:
dx * dy = det(M) * dr * dtheta
As for the limits, the original integral was x >= 0, y >= 0, which is
simply the first quadrant, the whole first quadrant, and nothing but
the first quadrant. In polar coordinates, that means the angle theta
is 0 to pi/2 (i.e., the first quadrant), and any positive r.
Michael
Ioannis
news:etpnd2$lar$1...@news.datemas.de...
[snip]
> Why? I can see that r^2=x^2+y^2, but I can't see why dx*dy = r*dr*dtheta. Do
> you mind explaining this step and the new bounds?
See the last section of "Change of Variables" at Wiki:
http://en.wikipedia.org/wiki/Change_of_variable
Your transformation is given by:
x = r*cos(theta)
y = r*sin(theta)
therefore the differentials transform as:
dx*dy = |J_F(r,theta)|*dr*dtheta, where J_F(r,theta) is the Jacobian of the
transformation and |.| its determinant.
In particular, |J_F(r,theta)| = r, as another poster showed, so:
dx*dy = r*dr*dtheta.
> Thanks!
> --
> Matt
--
I.N. Galidakis
http://ioannis.virtualcomposer2000.com/
Passerby
[snip]
>> >I^2 = integral theta=0 to pi/2 integral r = 0 to infinity e^(-r^2) * r
>> >dr dtheta
>>
>> Why? I can see that r^2=x^2+y^2, but I can't see why dx*dy = r*dr*dtheta. Do
>> you mind explaining this step and the new bounds?
Conceptually, one can think of an area element in polar coordinates as
dA = ds dr.
But ds = r dtheta (similar to the formula s = r theta for an arc
length of a circle)
So dA = (r dtheta) dr = r dr dtheta
The World Wide Wade
david....@gmail.com wrote:
> How do you calculate the value of an integral when the function
> doesn't have an antiderivative. E.g. in class we found that ｺ e^(-
> x^2) dx from 0 to ｰ is 1/2 × √π using a calculator, but how would one
> actually go about finding this?
Every continuous function on R has an antiderivative, including e^(-x^2).