definite integral.....

This is an improper integral called Gaussian integral:
To justify the improper double integrals and equating the two expressions, we begin with an approximating function:


圖片參考：http://upload.wikimedia.org/math/c/e/4/ce43a0a87dace80adc9d9f706eff9cea.png

so that the integral may be found by


圖片參考：http://upload.wikimedia.org/math/e/7/3/e73ae3961540cda4462fa517586d3c6b.png

Taking the square of I yields


圖片參考：http://upload.wikimedia.org/math/2/9/8/298c0579d012e0721a724502829aed48.png

Using Fubini's theorem, the above double integral can be seen as an area integral
圖片參考：http://upload.wikimedia.org/math/b/3/a/b3a311610e2638cc8b65c9e6c90aad47.png
, taken over a square with vertices {(−a, a), (a, a), (a, −a), (−a, −a)} on the xy-plane.
Since the exponential function is greater than 0 for all real numbers, it then follows that the integral taken over the square's incircle must be less than I(a)2, and similarly the integral taken over the square's circumcircle must be greater than I(a)2. The integrals over the two disks can easily be computed by switching from cartesian coordinates to polar coordinates:
圖片參考：http://upload.wikimedia.org/math/b/9/4/b94f1fd239d7676b1d229295c7fed0c9.png
:


圖片參考：http://upload.wikimedia.org/math/7/4/2/74223a2b03019242859ff2fb7d28e25e.png

Integrating,


圖片參考：http://upload.wikimedia.org/math/1/c/0/1c0e38b91b0177614244215bd8f84c79.png

By the squeezing principle, this gives the Gaussian integral


圖片參考：http://upload.wikimedia.org/math/1/c/f/1cf508d37b85da89fddbc6b3edb0d480.png