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Problem:True/False


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.


Helpdesk

  • What is this topic: Real Analysis
  • What are some of the associated concept: Monotonic increasing, Comparison Test, Bounded Integral
  • Book Suggestions: Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert