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When I'm learning convergence, my teacher just show me about condition to convergence or not. But I haven't meet a function that contain both trigonometric and normal polynomial. When I asked one of my friends, he tell me that using Taylor to developed $\cos x$, but I'm afraid that is not the good solution.

$$\int_0^\infty\frac{\cos^2(x)}{x^2+5x+11}dx$$

Thanks :)

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3 Answers 3

up vote 5 down vote accepted

This is one of those questions that looks far worse than it really is.

So we consider $\displaystyle\int_0^\infty\frac{\cos^2x}{x^2+5x+11}dx$

Note immediately that the denomonator $x^2 + 5x + 11 > 11$ over the domain, and the numerator is bounded above by $1$. So on any finite subinterval, this integral is bounded. More to the point, we might consider only $\displaystyle \int_1^\infty\frac{\cos^2x}{x^2+5x+11}dx$, as the integral from $0$ to $1$ is bounded.

But $\displaystyle \int_1^\infty\frac{\cos^2x}{x^2+5x+11}dx < \int_1^\infty\frac{1}{x^2+5x+11}dx < \int_1^\infty\frac{1}{x^2}dx$

And there we have it.

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Let $$I = \int_0^{\infty} \frac{\cos^2(x)}{x^2 + 5x+11} dx$$ The integrand is non-negative and since $\cos^2(x) \in [0,1]$, $\forall x \in \mathbb{R}$, we get that $$0 \leq I = \int_0^{\infty} \frac{\cos^2(x)}{x^2 + 5x+11} dx \leq \int_0^{\infty} \frac1{x^2 + 5x+11} dx = \int_0^{\infty} \frac{dx}{(x+5/2)^2 + 19/4}$$ Now make use of the identity $$\int \frac{dx}{x^2 + a^2} = \frac{\arctan (x/a)}a$$ and bound the integral $I$. Hence, we get that $$0 \leq I \leq \left. \frac{2}{\sqrt{19}} \arctan \left(\frac{2x+5}{\sqrt{19}} \right) \right \rvert_{0}^{\infty} = \frac{2}{\sqrt{19}} \left( \frac{\pi}{2} - \arctan \left(\frac{5}{\sqrt{19}}\right) \right) = \frac{\pi - 2 \arctan \left( \dfrac{5}{\sqrt{19}} \right)}{\sqrt{19}} \approx 0.328984$$

EDIT Another quick argument is along the lines of what mixedmath has done but there is no need to split the integral over two regions $$I = \int_0^{\infty} \frac{\cos^2(x)}{x^2 + 5x+11} dx \leq \int_0^{\infty} \frac1{x^2 + 5x+11} dx \leq \int_0^{\infty} \frac{dx}{(x+5/2)^2} = \left. \left(-\frac1{x+5/2} \right) \right \rvert_{0}^{\infty} = \frac25$$

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You can see that $\displaystyle \int_0^\infty \dfrac{\cos^2 x}{x^2 + 5x + 11} dx \le \int_0^\infty \dfrac{1}{x^2+5x+11}dx$ while the RHS expression converges, thus your integral is convergent too.

When it comes to use its Taylor series, after expanding, you can go on by proving that each term of your expansion is uniformly convergent to a function (there may be other way than this). After than, you can integrate each term of the series and the problem usually comes when you have to evaluate the value of new series.

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