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Regarding the integral $\int^\infty_0\frac{x^2+1-\cos x}{(x^2+1)x^p} dx$ where $p$ is a real number. When does it converge and why? I am guessing that the necessary and sufficient condition be $1

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    wolfram reports it converges for values larger than 2. for example, $p=e$ – 2017-01-24
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    It was because $\lim_{x\to0+}\frac{x^2+1-\cos x}{(x^2+1)x^{2.1}}=+\infty$ that I thought we need $p\le2$. – 2017-01-24

2 Answers 2

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I find that the easiest way to do these problems is to argue non-rigorously. All of the following $\approx$ signs can be made rigorous, but morally, this is the correct argument:

As $x \to \infty$, we see $x^2 + 1 - \cos(x) \approx x^2$ and $(x^2+1)x^p \approx x^{p+2}$. Thus $$\frac{x^2 + 1 - \cos(x)}{(x^{2}+1)x^p} \approx \frac{1}{x^p}.$$ Thus for convergence as $x \to \infty$, we need $p > 1$.

As $x \to 0^+$, we see $\cos(x) \approx 1$ so $x^2 + 1 - \cos(x) \approx x^2$ and similarly $(x^2+1)x^p \approx x^p$. Thus $$\frac{x^2 + 1 - \cos(x)}{(x^{2}+1)x^p} \approx x^{2-p}.$$ For convergence near $x = 0$, we need $2-p > -1$ which is the same as $p < 3$.

Thus the integral converges for $1 < p < 3$.

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    For values $x\to0$ and $x\to\infty$ the approximation $\cos x$ with a Taylor expansion is so meaningless. – 2017-01-24
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    What do you mean? – 2017-01-24
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    I mean $\cos x\approx1$ valid for $x\to0$ only. – 2017-01-24
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    So it is not that $x^2+1-\cos x\approx x^2$ which caused the first $\approx$? It was because for large $x$, $1-\cos x$ is bounded? – 2017-01-24
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    Of course $\cos(x) \approx 1$ is only valid near $x = 0$. I didn't use $\cos(x) \approx 1$ near $x = \infty$, instead I used that $\cos(x)$ is bounded as $x \to \infty$ so, for example, for large enough $x$: $$\frac{x^2 +1 - \cos(x)}{(x^2+1)x^p} \le \frac{x^2 + 2}{x^{p+2}} \le \frac{2x^2}{x^p} = \frac{2}{x^p}.$$ The first bound in the above line holds for all $x$ and the second when $x > \sqrt 2$. There was no use of $\cos(x) \approx 1$ here. I only used $\cos(x) \approx 1$ in the bound as $x \to 0$. – 2017-01-24
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Let $f(x)=\cos x-x^2-1+x^p$ on $[0,\infty)$. $f'(x)>0$ for $1f(0)=0$ this means $x^p>x^2+1-\cos x$ or $$\frac{x^2+1-\cos x}{x^p}<1$$ With the integral $$\int^\infty_0\frac{x^2+1-\cos x}{(x^2+1)x^p} dx<\int^\infty_0\frac{1}{x^2+1} dx=\arctan x\Big|^\infty_0=\frac{\pi}{2}$$ so the integral converge for $1

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    Did not you said that the condition is $1 – 2017-01-24