Question

Test the following improper integral for convergence :

Answer 1

\[\int\limits_{0}^{\infty} e ^{-x ^{3}} dx\]

Answer 2

i cant approach the solution even with support :
http://math.stackexchange.com/questions/96840/does-the-improper-integral-int-0-infty-e-x2dx-converge

Answer 3

Integrals are the continuous analogues of discrete series. What this means is that a comparison test (like the one you would use for an infinite series) will generally work for integrals as well.

As suggested in the link you posted, try comparing \(e^{-x^3}\) to something you CAN integrate, like \(e^{-x}\) (you can also compare to \(e^{-x^2}\), but the computation will be a bit more involved).

You know that for \(x>0\), \(e^{x}<e^{x^3}\), right? This means \(e^{-x}>e^{-x^3}\). The comparison test for series says if series A converges and A is larger than series B, then B must also converge. Furthermore, if A diverges and is smaller than B, B must also diverge.

This means
\[\int_0^\infty e^{-x^3}~dx~~\text{ will converge if }~~\int_0^\infty e^{-x}~dx~~\text{ converges.}\]