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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