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I have been trying to determine whether the following improper integral converges or diverges: $$\int_0^\infty \frac{\sin^4x}{x^2}dx$$

I have parted it into two terms. The first term: $$\int_1^\infty \frac{\sin^4x}{x^2}dx$$ converges (proven easily using the comparison test), but the second term: $$\int_0^1 \frac{\sin^4x}{x^2}dx$$ troubles me a lot. What could I do?

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  • $\begingroup$ Hint: $\frac{\sin^{4}(x)}{x^{2}}$ is continuous on $[0,1]$ $\endgroup$ – TheOscillator Jun 11 '15 at 13:13
  • $\begingroup$ A Taylor expansion of the numerator around $x=0$ will suffice. $\endgroup$ – Giuseppe Negro Jun 11 '15 at 13:14
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    $\begingroup$ you can bound the integral by $1/x^2$ on $[1, \infty]$. on $[0, 1]$ use $sin(x)\leq x$ $\endgroup$ – tired Jun 11 '15 at 13:15
  • $\begingroup$ @tired You are right, that works as well. Thanks. $\endgroup$ – user209217 Jun 11 '15 at 13:26
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For the second term, you can re-write your function as $\frac{\sin x}{x} \cdot \frac{\sin x}{x} \cdot \sin^2(x)$. Note that on $[0,1]$, this function is continuous (by the Fundamental Trig Limit you can extend the function to be defined as $f(0)=1$ at $x=0$). But then any continuous function on a closed interval is integrable, so $\int_0^1 \frac{\sin^4(x)}{x^2}$ converges. Hence the whole integral converges.

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  • $\begingroup$ Clever solution to the problem. Thanks for your answer! $\endgroup$ – user209217 Jun 11 '15 at 13:23
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The inequality $\left|\sin x\right|\leq\min(1,|x|)$ is enough to prove that the integral is converging.

Integration by parts and the well-known $\int_{0}^{+\infty}\frac{\sin x}{x}\,dx=\frac{\pi}{2}$ then give: $$\begin{eqnarray*} I = \int_{0}^{+\infty}\frac{\sin^4 x}{x^2}\,dx &=& \int_{0}^{+\infty}\frac{4\cos x \sin^3 x}{x}\,dx\\&=&\int_{0}^{+\infty}\frac{\sin(2x)-\frac{1}{2}\sin(4x)}{x}\,dx\\&=&\left(1-\frac{1}{2}\right)\int_{0}^{+\infty}\frac{\sin x}{x}\,dx=\color{red}{\frac{\pi}{4}}.\end{eqnarray*}$$

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