# Convergence of Improper Integral (TIFR 2013 problem 17)

Question:

True/ False?

The integral $\int_{0}^{\infty} e^{-x^5}dx$ is convergent.

Hint:

As x varies from 0 to “infinity”, $-x^5$ varies from 1 to “minus infinity”. Basically, the function is “rapidly decreasing”. We can hope that nothing goes wrong in the convergence.

Discussion:

Recall that $\int_{0}^{\infty} e^{-x}dx$ is convergent. (In fact the value of this integral is 1 which can be done by simple calculation). We try to use this for our comparison test. For $1\le x < \infty$   $e^{-x^5} \le e^{-x}$. This is because in this interval, $x^5 \ge x$ and because $e^x$ is an increasing function. So for $1\le x < \infty$, our worries are over.

$\int_{1}^{\infty} e^{-x^5} \le \int_{1}^{\infty} e^{-x}dx < \infty$.

Now we only need to check whether the integration is finite on $[0,1]$. The intergand is continuous and bounded by 1 (because it is monotonic decreasing and value at 0 is 1) on this interval, hence the integral is bounded. $\int_{0}^{1} e^{-x^5}dx \le \int_{0}^{1} 1dx = 1$.

Thus, $\int_{0}^{\infty} e^{-x^5}dx$ is convergent.