Help with a tricky improper integration problem

[stonecoldgen]

Show that:

∫(x^2)e^(-x^2)=(1/2)∫(e^(-x^2)

(both integrals are from 0 to ∞)



I don't even know if it's possible to integrate these functions, so it makes me think that what's interesting about the problem is that it only works with this integration limits.

I know improper integration, but don't know if prooving this "numerically" is the right path.

Thanks, I would appreciate any advice

 

[Curious3141]

Show that:

∫(x^2)e^(-x^2)=(1/2)∫(e^(-x^2)

(both integrals are from 0 to ∞)
I don't even know if it's possible to integrate these functions, so it makes me think that what's interesting about the problem is that it only works with this integration limits.

I know improper integration, but don't know if prooving this "numerically" is the right path.

Thanks, I would appreciate any advice

Hint:

$$\int_0^{\infty} x^2e^{-x^2}dx = \int_0^{\infty} x.(xe^{-x^2})dx$$

Integrate that by parts (use $u = x, dv = xe^{-x^2}dx$), then use L'Hopital's Rule to calculate the limit of one of the expressions.

 

[stonecoldgen]

Hint:

$$\int_0^{\infty} x^2e^{-x^2}dx = \int_0^{\infty} x.(xe^{-x^2})dx$$

Integrate that by parts (use $u = x, dv = xe^{-x^2}dx$), then use L'Hopital's Rule to calculate the limit of one of the expressions.

is this correct?

I integrated by pats (differentiating x and integrating x^2e^{-x^2} )

until I got to the point where (e^{-x^2})/2 (evaluated from o to infinity)=0

Using L'Hopital's Rule, I effectively got 0=0




Is this the way?

 

[Curious3141]

is this correct?

I integrated by pats (differentiating x and integrating x^2e^{-x^2} )

until I got to the point where (e^{-x^2})/2 (evaluated from o to infinity)=0

Using L'Hopital's Rule, I effectively got 0=0




Is this the way?

No, I think you messed up the integration by parts. You should end up with:

$$\int_0^{\infty} x^2e^{-x^2}dx = xe^{-x^2}|_0^{\infty} +\frac{1}{2}\int_0^{\infty}e^{-x^2}dx$$

and the first expression on the RHS vanishes when you take the limit by L' Hopital's.

 

[stonecoldgen]

No, I think you messed up the integration by parts. You should end up with:

$$\int_0^{\infty} x^2e^{-x^2}dx = xe^{-x^2}|_0^{\infty} +\frac{1}{2}\int_0^{\infty}e^{-x^2}dx$$

and the first expression on the RHS vanishes when you take the limit by L' Hopital's.

It's the same thing (maybe my english isn't that good, I don't know) but Thanks!

 

[Curious3141]

It's the same thing (maybe my english isn't that good, I don't know) but Thanks!

How is it the same thing? You said you got "0 = 0" which means nothing. You also said you got $\frac{1}{2}\int_0^{\infty}e^{-x^2}dx = 0$ which is clearly wrong.

Anyway, if you're confident you have the right solution, then it's fine.

 

[stonecoldgen]

What about this one? Show:

∫e^-x2=∫√(-lny)
(from 0 to ∞) (from 0 to 1)



I tried this one by solving for y in the right side of the equation, getting y=-e^x2

However, when checking the solution graphically, it's not right.

So what the hell am I supposed to do? I also wonder if the integration limits changed when expressed in y or x.


Thanks

 

[Millennial]

You should do a substitution to change the limits and get the log into play, then integrate by parts. Your previous result might help here.

 

[Curious3141]

What about this one? Show:

∫e^-x2=∫√(-lny)
(from 0 to ∞) (from 0 to 1)



I tried this one by solving for y in the right side of the equation, getting y=-e^x2

However, when checking the solution graphically, it's not right.

So what the hell am I supposed to do? I also wonder if the integration limits changed when expressed in y or x.


Thanks

Use this (your previous result):

$$\int_0^{\infty} x^2e^{-x^2}dx = \frac{1}{2}\int_0^{\infty}e^{-x^2}dx$$

and make the substitution $e^{-x^2} = y$ on the LHS. The result follows very quickly.