 TIFR 2013 Problem 17 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India’s premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
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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