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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.
The image is a front cover of a book named Introduction to Real Analysis by R.G. Bartle, D.R. Sherbert. This book is very useful for the preparation of TIFR Entrance.

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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