ò-¥¥ e-x2 dx
By Anonymous on Saturday, March 24,
2001 - 01:17 pm :
I have been looking at a way of evaluating ò-¥¥ e-x2 dx. However, I am getting stuck when it comes to using double integration in
the polar plane. Can anyone help?
By Kerwin Hui (Kwkh2) on Saturday,
March 24, 2001 - 01:54 pm :
Amars,
Let I=ò-¥¥ e-x2 dx
Then,
I2=ò-¥¥ e-x2 dxò-¥¥ e-y2 dy =ò-¥¥ò-¥¥e-x2e-y2 dy dx =ò02pò0¥e-r2r dr dq =p
so
Hence
Kerwin
By Emma McCaughan (Emma) on Monday,
March 26, 2001 - 09:50 am :
For another explanation, see Integral
of Gaussian function .
By Amars Birdi (P2769) on Wednesday,
March 28, 2001 - 05:05 pm :
Thanks for your replies, but I still can't see why, when converting from
cartesian to polar co-ordinates, dx dy=r dr dq. This has something to
do with areas - but what?
-Thanks!
By Amars Birdi (P2769) on Wednesday,
March 28, 2001 - 05:23 pm :
Also, why is it that when you multiply two separate integrals (one with
respect to y and the other with respect to x) you are allowed to end up
with an integral of an integral? More to the point, when you integrate
z=e-x2 this is an integral with respect to x, when you integrate
(another variable) =e-y2 this is an integral with respect to y. What
does it MEAN to multiply these two integrals - are we multiplying the
numerical values of each integral OR are we multiplying two separate
integrals in separate variables (I guess this is connected with what a
double integral is)
By David Loeffler (P865) on Wednesday,
March 28, 2001 - 08:55 pm :
Amars,
As regards the last question, we need a fundamental result called
Fubini's theorem, which states that the multiple integral
òòF(x,y) dx dy
is equal to the ''iterated integral''
ò( òF(x,y)dx ) dy
where we first evaluate the integral with respect to x, treating y as a
parameter, then integrate this with respect to y. (I've written this for two
variables but it extends to any number in the obvious manner.)
Of course if F(x,y) = f(x) g(y) then this becomes the product of the two
integrals òf(x)dx and òg(y)dy, which is the result we need to
evaluate the e-x2 integral.
David
By Michael Doré (Md285) on Wednesday, March 28,
2001 - 10:23 pm :
Here is an outline of another (non-standard) way
of approaching the original problem. It is straight off our Differential
Equations Sheet 1.
Let:
f(x)=(ò0x e-t2 dt)2 g(x)=ò01 e-x2(t2+1)/(1+t2)dt
Show that f(x)+g(x)=p/4. (Hint: differentiate the left hand side.)
By letting x®¥, show that
.
This is equivalent (by symmetry) to the required result.
By Dave Sheridan (Dms22) on Thursday,
March 29, 2001 - 01:26 pm :
Another way to view the
integral-multiplied-by-an-integral is as follows.
Suppose c is a constant. Then the integral of cf(x) is the same
as c times the integral of f(x). I think we're all pretty happy
with that.
Well, what if c isn't a constant exactly but doesn't depend on x?
For example, what's the integral of f(x)g(t) dx? Well, as far as
x can see, t isn't anything to do with x so g(t) is a constant
as x varies . So the integral is
equal to g(t) times the integral of f(x)dx.
Finally, what's the integral of (the integral of g(t)dt)f(x)dx?
Well, just as above we have that the integral of g(t)dt doesn't
change if we change x, so it's a constant from the point of view
of the integral dx. So we can conclude, just as above, that the
integral is equal to:
(the integral of g(t)dt)(the integral of f(x)dx).
This argument allows us to swap things around easily, but is
non-rigorous. Hopefully it makes sense though. Fubini's theorem
states that we may interchange the order of integration, even if
we can't factorise the integrals, so long as the integral itself
is finite. To prove this properly requires a lot of complicated
analysis, so I won't go into that here.
As for the confusion between multiplying two integrals together,
the entity (integral of f(x)dx) is simply a number, so we can
multiply integrals together as if we were simply multiplying two
numbers.
-Dave
By Brad Rodgers (P1930) on Thursday,
April 26, 2001 - 01:42 am :
Michael, I can't seem to figure how you can differentiate the
LHS and end up with (presumably) 0. Upon differentiation, I
get
4e-x2ò0x e-t2 dt
Have I missed something, or is there another method I need to
use?
Thanks,
Brad
By Brad Rodgers (P1930) on Thursday,
April 26, 2001 - 01:46 am :
Unless, of course, as I've just noticed, what one needs to do
is let x tend to infinity after differentiation, then integrate.
But is this allowed? If so, what allows us to do that?
Thanks,
Brad
By Kerwin Hui (Kwkh2) on Thursday,
April 26, 2001 - 12:58 pm :
OK, let's do this step-by-step.
It is easy to see that
|
f ' (x)= |
d
dx
|
|
æ
è
|
ó
õ
|
x
0
|
exp(-t2)dt |
ö
ø
|
2
|
=2 |
é
ë
|
ó
õ
|
x
0
|
exp(-t2) dt |
ù
û
|
exp(-x2) |
|
|
|
|
|
|
ó
õ
|
1
0
|
|
¶
¶x
|
|
exp[-x2(t2+1)]
1+t2
|
dt |
| |
|
| -2x |
ó
õ
|
1
0
|
exp[-x2(t2+1)]dt=-2 exp(-x2) |
ó
õ
|
x
0
|
exp(-t¢2) dt¢ |
|
|
and so (f ' +g ' )(x) º 0 for all x.
Kerwin