What is the Solution to the Integral of e^-2x^2?

[Moneer81]

Hello,

the following integral confused me a little bit:

[tex] \int_{0}^{\infty} e ^ {-2.x^2} dx [/tex]

The answer is [tex] \sqrt \frac{\pi}{8} [/tex]

and I have no idea where this answer came from. Do I start by saying let

[tex] u = e^ {2x^2} [/tex] ?

 

[HallsofIvy]

No, unfortunately the anti-derivative of [tex]e^{-2x^2}[/tex] cannot be written in terms of elementary functions. Where did you see that integral and how do you know the answer is [tex] \sqrt \frac{\pi}{8} [/tex] (which it is)? I ask that since it would help to know at what level this is- what kind of mathematics are you doing? I cannot imagine that you were told to integrate that in a basic calculus course unless you were given additional information. For example, if you already knew that [tex]\int_0^\infty e^{-x^2}dx= \frac{\sqrt{\pi}}{2}[/tex] it would be simple.

 

[Moneer81]

well I've done Calculus II a long time ago and I thought that an exponentional raised to a power (that is a function) should be relatively simple to integrate. Care to tell me more about your suggestion? I am pretty sure I'll understand what you say because I've had quite a bit ofmath but it has been a long time ago so I'll remember it.

and the answer was in the book but I am interested in knowing how we got it.

 

[HallsofIvy]

Okay, take a deep breath, and fasten your seat belt!

Let [tex]I= \int_{0}^{\infty} e ^ {-2.x^2} dx [/tex].
Of course we also can write [tex]I= \int_{0}^{\infty} e ^ {-2.y^2} dy [/tex] since I does not depend on the "dummy" variable of integration.

Multiplying those together, [tex]I^2= \left(\int_{0}^{\infty} e ^ {-2.x^2} dx\right)\left( \int_{0}^{\infty} e ^ {-2.y^2} dy\right)[/tex].

Since one integral involves on x and the other only y, and the neither limit of integration depends on the other variable, we can combine them into a double integral:
[tex]I^2= \int_{0}^{\infty} \int_{0}^{\infty} \left(e ^ {-2.x^2}\right)\left(e^{-2y^2}\right) dxdy=\int_{0}^{\infty} \int_{0}^{\infty} \left(e ^ {-2(x^2+ y^2)}\right)dxdy [/tex]

We can interpret that as integrating over the first quadrant (x and y both positive) and convert to polar coordinates.
[tex]e^{-2(x^2+ y^2)}[/tex] is easy: its [itex]e^{-2r^2}[/itex] in polar coordinates. The r-integral is from 0 to [itex]\infty[/itex] and the [itex]\theta[/itex] integral is from 0 to [itex]\frac{\pi}{2}[/itex] of course. Finally, the "differential of area" in polar coordinates is [itex]rdrd\theta[/itex].
We have
[tex]I^2= \int_0^{\frac{\pi}{2}}\int_0^\infty e^{-2r^2}rdrd\theta[/tex].

Of course, the [itex]\theta[/itex] integral just give [itex]\frac{\pi}{2}[/itex] so
[tex]I^2= \frac{\pi}{2}\int_0^\infty e^{-2r^2} rdr[/tex]

The crucial point is that we now have "rdr"! Make the substitution u= 2r². Then du= 4 rdr. or rdr= (1/4)du. Now we have
[tex]I^2= \frac{\pi}{8}\int_0^\infty e^{-u} du[/tex].

I'll leave it to you to finish that.

 

[benorin]

Ivy, I have read (and posted myself) the above post several times. Do you think it ought to find itself in an FAQ (pinned to the top of the forum listing) amongst posts such as: "Why 0.999...=1", "No, you didn't just prove that 0=1", "A review of Partial Fraction Decomposition", , ...

-Ben

 

[Moneer81]

HallsofIvy,

Thank you very much for your help with this problem, I really appreciate it. I am trying to remember where I did something like this in the past. Do you know what do I need to study to be able to become more comfortable with such integrals?

thanks again