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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.