[數撚圍爐區] 農曆都有聖誕? (25)
終極連登數學白痴
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Thats not true. The second integral does not exist cuz it equals \lim_{a -> -\infty} a^2 which does not exist.
Oh no, it’s zero i misread the question im sorry lol
其實我打多左個負號不過唔影響個結果 
That brings the interesting question: how should one define improper integrals like \int_{-infty}^{\infty} f(x)dx?
a classical way is the following:
for a real number d, define
\int_d^infinity f(x) dx = \lim_{a -> infinity} \int_d^a f(x) dx
and define \int_{-infinity}^d f(x) dx similarly,
then, if there is a "cutting point" c such that
\int_c^infinity f(x) dx and \int_{-infinity}^c f(x) dx are finite numbers
we define \int_R f(x) dx be the sum of them,
and then you can show that if such point exists, then any point is also a "cutting point" and the value of the sum remains unchanged.
for a real number d, define
\int_d^infinity f(x) dx = \lim_{a -> infinity} \int_d^a f(x) dx
and define \int_{-infinity}^d f(x) dx similarly,
then, if there is a "cutting point" c such that
\int_c^infinity f(x) dx and \int_{-infinity}^c f(x) dx are finite numbers
we define \int_R f(x) dx be the sum of them,
and then you can show that if such point exists, then any point is also a "cutting point" and the value of the sum remains unchanged.
