Question

a>0. integrate from 0 to infinity of e^(-ax^2)

Answer 1

Uh, I think it's \(\large \sqrt{\frac{\pi}{4a}}\).

Answer 2

i cannot understand it, how we know that

Answer 3

I thought you wouldn't.

Do you know \(\large \int_0^{\infty}e^{-x^2}\mathrm{d}x\) ?

Answer 4

no.

Answer 5

Ok, so I will show you how to find that, and then you can find your answer easily.

Answer 6

should have put 2 outside the root

Answer 7

So consider the following integral:\[\int_0^{\infty}\int_0^{\infty} e^{-x^2-y^2}dy~dx=\int_0^{\infty}\int_0^{\infty} e^{-x^2}e^{-y^2}dy~dx=\int_0^{\infty}e^{-x^2}\int_0^{\infty} e^{-y^2}dy~dx\]\[=\left (\int_0^{\infty}e^{-x^2}dx\right ) \left (\int_0^{\infty} e^{-y^2}dy\right ) =\left (\int_0^{\infty}e^{-x^2}dx\right )^2\]

Answer 8

in second last step, do we consider y = x?

Answer 9

@eamier Well, x and y are both "dummy variables". The value of a definite integral does NOT depend on what variable we are using. It is simply a number. It depends on the function and the region of integration.

So, if we find the volume under the surface e^(-x^2-y^2) in the first quadrant, we have found the square of our desired integral. (I hope you understand volume integrals. If not, look in your textbook or whatevs.)

Now \(-x^2-y^2=-\left ( x^2+y^2\right ) =-\left ( \sqrt{x^2+y^2}\right ) ^2=-r^2\), where r is the distance from the origin, used in polar coordinates.

Answer 10

So the surface depends only on r. Not on Θ. You should realize that this means the surface is a surface of revolution. You probably studied such surfaces in Calc 1. What we have is a function z=f(r) rotated around the z axis.

Answer 11

But we only need one fourth of the entire volume.

Answer 12

Do you think you can get it from there?

Answer 13

this really has nothing to do with when you integrate e, divide by the derivative of it power..i really cant imagine it. but how we get the pi

Answer 14

Or take: \[\int_0^{\pi/2}\int_0^{\infty}e^{-r^2}r~dr~d\theta\]This is a simple polar integral which represents the volume under \(e^{-x^2-y^2}\) in the first quadrant.

I know it may seem strange, but the math is flawless. You get the pi because multiplying \(e^{-x^2}\) with \(e^{-y^2}\) just so happens to give you \(e^{-r^2}\), and anything to do with polar coordinates usually has to do with pi. You will find that the above integral is equal to \(\frac{\pi}{4}\).

Once you know that integral, you can use simple substitutions to find your original integral.

Do you need help with the above integral?

Answer 15

i tried but cant do even this part\[\int\limits_{0}^{\infty}e ^{-r ^{2}}rdr\].

Answer 16

let r^2 = u
2rdr = du

Answer 17

done. tq