← 2022 Paper 2

UPSC 2022 Maths Optional Paper 2 Q1c — 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 $\displaystyle\int_0^\infty\dfrac{\cos x}{1+x^2}\,dx$.

Technique

Comparison test with $1/(1+x^2)$.

Solution

Strategy. Use absolute convergence: if $\int|\cos x/(1+x^2)|\,dx$ converges, so does the original.

Step 1 — Bound the integrand

$|\cos x|\le 1$, so $\left|\dfrac{\cos x}{1+x^2}\right|\le\dfrac{1}{1+x^2}$.

Step 2 — $\int_0^\infty\dfrac{dx}{1+x^2}$ converges

$\int_0^\infty\dfrac{dx}{1+x^2}=[\arctan x]_0^\infty=\pi/2-0=\pi/2<\infty$.

So $\int_0^\infty\dfrac{dx}{1+x^2}$ converges.

Step 3 — Apply comparison

By comparison, $\int_0^\infty\left|\dfrac{\cos x}{1+x^2}\right|dx\le\int_0^\infty\dfrac{dx}{1+x^2}=\pi/2$.

So $\int_0^\infty\dfrac{\cos x}{1+x^2}dx$ converges absolutely, hence converges.

Answer