Testing for convergence - $\int_{0}^{\infty} \frac{x^4}{e^{\sqrt{x}}} dx$

Question

How do I go about testing for the convergence of the following improper integral: $$\int_{0}^{+\infty} \frac{x^4}{e^{\sqrt{x}}} dx$$

I don't suppose it's possible (or practical) to evaluate its antiderivative. I tried testing around for an asymptotically equivalent function in the neighbourhood of $+\infty$, I've also tried to look for a fitting function to make the comparison test; both to no avail.

Answer 1

You may split your integral in two, one from $0$ to $1$ and other from $1$ to $\infty$. It is obvious the first one is finite due to continuity of the integrand.

Furthermore, when $x\to\infty$, $\dfrac{x^4}{e^{\sqrt{x}}}$ is way smaller than, for example, $\dfrac{1}{x^2}$, as the exponential 'takes over'.

More formally, $$\lim_{x\to\infty}\dfrac{\dfrac{x^4}{e^{\sqrt{x}}}}{\dfrac{1}{x^2}}=\lim_{x\to\infty}\dfrac{x^6}{e^\sqrt{x}}=0$$

Then, by comparison, as $$\int_{1}^{\infty} \dfrac{1}{x^2}<\infty,$$

we have that $$\int_{1}^{\infty} \dfrac{x^4}{e^\sqrt{x}}<\infty$$ and so the original integral, from $0$ to $\infty$, converges.

Answer 2

Hint The substitution $x = u^2, dx = 2 u \,du$, transforms the integral to $$2 \int_0^\infty u^9 e^{-u} \,du.$$

So, in fact, the integral has value $2 \cdot 9!$.

Answer 3

We can apply infinite series. Namely $$\int\limits_{n^2}^{(n+1)^2 }{x^4\over e^{\sqrt{x}}}\,dx\le {(n+1)^8\over e^n}(2n+1)$$ By the Cauchy $n$-th root test the series $\sum{(n+1)^8\over e^n}$ is convergent, so is the integral in question.