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

Question

Christopher Allers · Accepted Answer

#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.

Isabella Ackerman · Answer

undefined To determine whether the integral of $ e^{-x^2} $ from $ 0 $ to $ \infty $ converges or diverges, we use the integral test for convergence. This involves comparing the function's integral to a known convergent or divergent integral. However, the integral of $ e^{-x^2} $ does not have a elementary antiderivative, so we cannot integrate it directly. Consequently, we employ a different approach, considering its behavior within the integral.

The function $ e^{-x^2} $ is always positive, continuous, and decreasing on the interval $[0, \infty)$. Thus, we can't use the integral test directly. However, we can compare it to the Gaussian integral $ \int_{0}^{\infty} e^{-x^2} \, dx $, which is a well-known convergent integral. Since $ e^{-x^2} \leq e^{-x} $ for all $ x \geq 1 $, and $ e^{-x} $ is integrable on $[1, \infty)$, we can conclude that $ e^{-x^2} $ is also integrable on $[1, \infty)$, and therefore on $[0, \infty)$. This implies that the integral $ \int_{0}^{\infty} e^{-x^2} \, dx $ converges.

Thus, by comparison with the convergent integral $ \int_{0}^{\infty} e^{-x^2} \, dx $, we can conclude that $ \int_{0}^{\infty} e^{-x^2} \, dx $ also converges.