How do you use the integral test to determine whether #int e^(-x^2)# converges or diverges from #[0,oo)#?

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Answer 1 · 2017-02-28T00:46:46

#int_0^oo e^(-x^2)dx# is convergent.

As #f(x) = e^(-x^2)# is positive, strictly decreasing and infinitesimal for #x->oo# the convergence of the integral:

#int_0^oo e^(-x^2)dx#

is equivalent to the convergence of the sum:

#(1) sum_(n=0)^oo e^(-n^2)#

based on the integral test theorem.

We can demonstrate that the series #(1)# is convergent based on the root test:

#lim_(n->oo) abs(a_n)^(1/n) = lim_(n->oo)( e^(-n^2))^(1/n) = lim_(n->oo) e^((-n^2)/n) = lim_(n->oo) e^-n = 0#

so the integral must also be convergent.