UPSC 2014 Maths Optional Paper 2 Q1b — Step-by-Step Solution

10 marks · Section A

Improper integrals (analysis perspective) · Real Analysis · asked 4× in 14 yrs · Read the full method →

Question

Test the convergence of the improper integral $\displaystyle\int_1^\infty\frac{dx}{x^2(1+e^{-x})}$ .

Comparison test for improper integrals; bound $0<\frac{1}{1+e^{-x}}<1$ .

Strategy. Bound the integrand by a simpler comparable expression; use the comparison test.

For $x\ge 1$ : $0<e^{-x}\le e^{-1}<1$ , so

1<1+e^{-x}\le 1+e^{-1}.

Equivalently, $\dfrac{1}{1+e^{-1}}\le\dfrac{1}{1+e^{-x}}<1$ .

In particular, $0<\dfrac{1}{1+e^{-x}}<1$ for all $x\ge 1$ , giving:

0<\frac{1}{x^2(1+e^{-x})}<\frac{1}{x^2}.

The integral $\displaystyle\int_1^\infty\frac{dx}{x^2}=\left[-\frac{1}{x}\right]_1^\infty=1$ is convergent.

By the comparison test, the smaller positive integrand $\dfrac{1}{x^2(1+e^{-x})}$ also gives a convergent integral.

\boxed{\;\text{The integral converges.}\;}