Does the following integral have a finite value? How to compute it? $$\int_{0}^{+\infty} e^{-x^k}\mathrm{d} x$$ where $k$ is given and $0
Compute $\int_{0}^{+\infty} e^{-y}y^a\mathrm{d} y$
7
$\begingroup$
real-analysis
integration
-
1Take a look at http://en.wikipedia.org/wiki/Gamma_function – 2012-09-11
-
1That looks like a [gamma function](http://en.wikipedia.org/wiki/Gamma_function). – 2012-09-11
-
1@Tunococ: 11 seconds :) – 2012-09-11
-
3Existence does not require knowing about the Gamma function. Our integrand behaves nicely near $0$, and in the long run decays (far) more rapidly than $1/x^2$. – 2012-09-11
-
0@AndréNicolas: I got your meaning that once the integrand decays faster than $1/x^2$ and then the integral has a finite value. Is it correct in the long run the integrand decays more rapidly than any $1/x^k$ with $k>0$? – 2012-09-11
-
0@Shiyu: Yes, it is correct. For the comparison, I picked the smallest *integer* $k$ that does the job, for no special reason. For sure we want to use a $k\gt 1$. – 2012-09-11
1 Answers
0
I am formalizing the comments:
$$\Gamma(a)=\int_0^\infty e^{-y}y^{a-1}\text{ d}y$$
$$\Gamma(a+1)=\int_0^\infty e^{-y}y^a\text{ d}y$$
Convergence:
So we know that $\Gamma(x)$ converges (for at least $x\geqslant0$)
because it converges significantly faster than $x^{-2}$ for arbitrarily large $x$ values.