Get the solution to "Show that the integral is divergent ∫0∞dx1+x2sin2x"

Aaron Coleman

Answered question

2022-05-03

Show that the integral is divergent

\int_{0}^{\infty} \frac{d x}{1 + x^{2} \sin^{2} x}

ritmesysv

Beginner2022-05-04Added 12 answers

We have that

\int_{0}^{\infty} \frac{d x}{1 + x^{2} \sin^{2} x} = \sum_{n = 0}^{\infty} \int_{0}^{π} \frac{d x}{1 + {(x + n π)}^{2} \sin^{2} (x + n π)}

\geq \sum_{n = 0}^{\infty} \int_{0}^{\frac{π}{2}} \frac{2 d x}{1 + π^{2} {(1 + n)}^{2} \sin^{2} (x)}

= \sum_{n = 0}^{\infty} \frac{π}{\sqrt{1 + π^{2} {(n + 1)}^{2}}}

\geq \frac{π}{\sqrt{1 + π^{2}}} \sum_{n = 0}^{\infty} \frac{1}{n + 1} = + \infty

where we used the fact that

\int \frac{d x}{1 + a^{2} \sin^{2} (x)} = \frac{\arctan (\sqrt{1 + a^{2}} \tan (x))}{\sqrt{1 + a^{2}}} + C

P.S. We can also use the inequality

\sin^{2} (x) \leq x^{2}

and evaluate

\int_{0}^{\frac{π}{2}} \frac{2 d x}{1 + π^{2} {(1 + n)}^{2} x^{2}} = \frac{2 \arctan ((n + 1) π^{2} / 2)}{π (n + 1)}

eslasadanv3

Beginner2022-05-05Added 20 answers

As an alternative to first answer, you may consider that for any

n \in N^{+}

the function

x^{2} \sin^{2} x

is bounded by

2 π^{2}

on the interval

I_{n} = [π n - \frac{1}{n}, π n + \frac{1}{n}]

. Since the integrand function is non-negative and these intervals are disjoint,

\int_{0}^{+ \infty} \frac{d x}{1 + x^{2} \sin^{2} x} \geq \int_{\cup I_{n}} \frac{d x}{1 + 2 π^{2}} \geq \frac{1}{1 + 2 π^{2}} \sum_{n \geq 1} | I_{n} | = + \infty

since

| I_{n} | = \frac{2}{n}

and the harmonic series is divergent.

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