NationStates Jolt Archive

Since people are posting math problems, here's a fun one that I remember from college. I wonder if anyone can get it?

The problem is to calculate the following integral:

e^(-x^2)dx, x ranges from 0 to inf.

The answer is (sqrt pi)/2. But you have to figure out why! Have fun!

Isn't this a second year college calculus problem?

Some fun calculus.

That ranges up with world peace and microsoft works as the best Oxymoron ever...

well when I start calculus I'll try and give you an answer :p
and I vote microsoft works as best oxymoron :D

Calculus? Oh fuck off. It's bad enough I have to do it at uni every bloody working day! Get it off the forums!

*sets fire to the thread after stealing everyone's pants*

Since people are posting math problems, here's a fun one that I remember from college. I wonder if anyone can get it?

The problem is to calculate the following integral:

e^(-x^2)dx, x ranges from 0 to inf.

The answer is (sqrt pi)/2. But you have to figure out why! Have fun!

this is a fun one indeed and we have seen 2 different ways of solving the Poisson integral:

the area G is R**2 and f(x,y) = e^-(x**2+y**2)

the row Gn = B(0,n) (a circle with 0 as the middle and n as the radius) fills R**2

in circle coordinats we get:

int(f(x,y),dx,dy) over Gn = int(int(r*exp(-r**2),r=0..n),theta=0..2*Pi)

= Pi*(1-exp(-n**2))

with n-> infinity, Pi*(1-exp(-n**2)) -> Pi

thus: int(int(exp(-(x**2+y**2))*dx)*dy) over R**2 = Pi

now take the row Hn of closed intervals Hn = [-n,n] X [-n,n], this clearly fills R**2.

int(int(f(x,y)*dx)*dy) over Hn = int(int(exp(-(x**2+y**2)),dy=-n..n),dx=-n..n)
= [int(exp(-x**2),dx=-n..n)] * [int(exp(-y**2),dy=-n..n)]
= [int(exp(-x**2),dx=-n..n)]**2

the limit of this one has to be Pi so we get that:

int(exp(-x**2),dx=-infinity..infinity) = sqrt(Pi)


The second method:

we have:
F: x->F(x) = int(exp(-x**2*(1+t**2))/(1+t**2),dt=0..1)
the functions
f: (x,t)->exp(-x**2*(1+t**2))/(1+t**2)
and Dxf: (x,t) -> -2*x*exp(-x**2*(1+t**2))
are both continuous in R**2
this allows us to say that
F'(x) = -2*x*int(exp(-x**2*(1+t**2)),dt=0..1)

substitute x*t = s and we receive:
F'(x) = -2*exp(-x**2) * int(exp(-s**2),ds=0..x)

take the function:
phi: x->phi(x) = int(exp(-s**2),ds=0..x)

now the Poisson integral is limit(phi(x), x->infinity)
and phi'(x) = exp(-x**2)

thus F'(x) = -2*phi'(x)*phi(x)
integration from 0 to x gives:

F(x) - F(0) = phi(0)**2 - phi(x)**2
we find that phi(0) = 0
and that F(0) = Pi/4

and we get:

phi(x)**2 = Pi/4-F(x)

because 0<= exp(-x**2*(1+t**2))/(1+t**2)<=exp(-x**2), for all t in [0,1],x in R
we get by integration over t that 0<=F(x)<=exp(-x**2)
and thus limit(F(x),x->infinity) = 0

we find:

limit(phi(x),x-> infinity) = sqrt(F(0)) = sqrt(Pi)/2

this is a fun one indeed

Glad someone agrees with me!

And I don't think I need to tell anyone that those are both fine solutions. :D