This is the legendary type that cannot be accumulated, so don't bother to integrate it.

Of course, if this problem is a generalized integral, you can have the following solution. The following answers come from the postgraduate entrance examination.

Method 1: The strict method is not very strict, but it is in the advanced mathematics textbook of Tongji University (the polar coordinate part of the second volume of double integral).

Let u = ∫ [-∞, +∞] e (-t 2) dt.

Square of both sides: (the integral limit is omitted below)

U 2 = ∫ e (-t 2) dt * ∫ e (-t 2) dt can change the integral variable at will because of the integral.

= ∫ e (-x 2) dx * ∫ e (-y 2) dy thus becomes a double integral.

= ∫ ∫ e (-(x 2+y 2)) dxdy integral area is x 2+y 2 = r 2r-> +∞

Polar coordinate replacement

=∫∫ e^(-r^2)*rdrdθ

=∫[0->； 2π]∫[0->； R] e (-r 2) * rdrd θ and then r-> +∞ take the limit.

= 2π*( 1/2)∫[0->； e^(-r^2)d (r^2)

= π [1-e (-r 2)], and then r-> +∞ takes the limit.

=π

So u 2 = π, so u=√π.

The inaccuracy is that when it is converted into a double integral, it should not be a circular area, but a rectangular area. There is such a treatment method in the book, and the rectangular area is sandwiched between two circular areas by pinching criterion to solve this problem.

Method 2: Match with the standard normal distribution in probability theory.

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