When does this improper Riemann integral converge?

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When $n$ is odd, how can we show that $$\int_1^{\infty} \ln\left(1+{{(\sin x)^n}\over{x^c}}\right)dx$$ exists as a Riemann integral if and only if $c \geq { 1 \over 2}$?

For the general techs used in such problems:
(1) When $x$ is small we have $\ln(1+x)=x$.
(2) Divide the interval $[1, \infty)$ into intervals $[1,2 \pi) \cup \left(\bigcup_{k=1}^{\infty} [2k\pi, (2k+1)\pi)\right) $ may work.

However, I cannot get $c \geq { 1 \over 2}$.

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The integral doesn't converge if $n=1,c=1/2.$ Proof:

$$\ln (1+ (\sin x)/\sqrt x))= (\sin x)/\sqrt x -\sin^2 x/(2x) + O(1/x^{3/2})$$

as $x\to \infty.$ The integral of $(\sin x)/\sqrt x$ converges by Dirichet's test. The integral of $-\sin^2 x/(2x)$ diverges to $-\infty.$ The integral of the $1/x^{3/2}$ term converges (absolutely). This adds up to divergence to $-\infty.$