2

$${I=\int^{\infty}_{0}\frac{\cos^{2} x}{x^2}\;dx=\infty}$$

Attempt:

$$\begin{align}&= \int^{\infty}_{0}\frac{1- \sin^{2} x}{x^2}\, dx \tag1 \\[8pt] &= \infty-\int^{\infty}_{0}\frac{\sin^{2} x}{x^2}\, dx \tag2 \\[8pt] &= \infty-\frac{\pi}{2} \tag3 \end{align}$$

Hence, $I$ is divergent.

Blue
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User1234
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2 Answers2

5

The reasoning in your question is almost correct. I prefer to avoid using $\infty$ like a real number, so I would proceed as follows.

If $$ \int_0^\infty\frac{\cos^2(x)}{x^2}\,\mathrm{d}x $$ converged, then $$ \int_0^\infty\frac{\cos^2(x)}{x^2}\,\mathrm{d}x+\underbrace{\int_0^\infty\frac{\sin^2(x)}{x^2}\,\mathrm{d}x}_{\text{convergent}}=\int_0^\infty\frac1{x^2}\,\mathrm{d}x $$ would also converge, but the latter integral diverges. Therefore, your original integral diverges.

robjohn
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  • Since the first integral above is convergent (it equals $\frac\pi2\log(\pi)$), either both of the latter integrals converge or both diverge (that is, if one converges, the other converges because it is the sum of two convergent integrals). In this example, as in your question above, I am only using that the sum (or equivalently, difference) of two convergent integrals is convergent. – robjohn Jun 05 '15 at 18:30
  • To evaluate the integral in your previous comment, I broke it into the limit of two proper integrals: $$\begin{align} \int_a^b\frac{\tan^{-1}(\pi x)-\tan^{-1}(x)}x,\mathrm{d}x &=\int_a^b\frac{\tan^{-1}(\pi x)}x,\mathrm{d}x-\int_a^b\frac{\tan^{-1}(x)}x,\mathrm{d}x\ &=\int_{\pi a}^{\pi b}\frac{\tan^{-1}(x)}x,\mathrm{d}x-\int_a^b\frac{\tan^{-1}(x)}x,\mathrm{d}x\ &=\int_b^{\pi b}\frac{\tan^{-1}(x)}x,\mathrm{d}x-\int_a^{\pi a}\frac{\tan^{-1}(x)}x,\mathrm{d}x \end{align}$$ as $b\to\infty$, the integral over $[b,\pi b]\to\frac\pi2\log(\pi)$, and as $a\to0$, the integral over $[a,\pi a]\to0$. – robjohn Jun 05 '15 at 18:42
  • No, I never split a convergent integral into two divergent integrals. I used that the sum (or difference) of two convergent integrals is convergent. If you are talking about the evaluation of the integral in my previous comment, none of the integrals is divergent since they are all integrals of finite functions over finite intervals. The infinite intervals are evaluated as limits. – robjohn Jun 05 '15 at 19:28
  • The first step was to split the integral of a difference into the difference of the integrals. Everything is good since all the functions are finite and the intervals are finite. The second step was to change variables in the first integral $x\mapsto x/\pi$. The third step was to cancel the difference of two identical integrals $\left(\int_b^{\pi b}\frac{\tan^{-1}(x)}{x},\mathrm{d}x+\int_{\pi a}^b\frac{\tan^{-1}(x)}{x},\mathrm{d}x\right)-\left(\int_{\pi a}^b\frac{\tan^{-1}(x)}{x},\mathrm{d}x+\int_a^{\pi a}\frac{\tan^{-1}(x)}{x},\mathrm{d}x\right)$ – robjohn Jun 05 '15 at 19:32
  • No. A convergent integral plus a convergent integral is convergent. An integral is convergent if and only if its negative is convergent. From these two facts we can deduce that if $A$ is convergent and $B$ is not convergent, then $A+B$ is not convergent. That is because if $A+B$ were convergent, then $B=(A+B)+(-A)$ would be convergent, which it is not. When you ask "can we write a convergent integral as a difference of two divergent integrals?" you are asking "can we make something sensible from two bits of nonsense?" – robjohn Jun 07 '15 at 01:26
2

This looks a little better, I'd say.

$$\int_0^\infty\frac{\cos^2x}{x^2}dx>\int_0^{\pi/4}\frac{\cos^2x}{x^2}dx\ge\frac12\int_0^{\pi/4}\frac{dx}{x^2}=\infty$$

ajotatxe
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  • First: Since $\cos^2x/x^2$ is positive, the integral is greater than the integral of the same function over a smaller interval. Second: for $x\in[0,\pi/4]$, $\cos^2 x\ge1/2$. Third, the integral of $x^{-n}$ over $[0,k]$ diverges for $n\ge1$ and $k>0$. – ajotatxe Jun 05 '15 at 17:44
  • @BetterWorld: where do you split an integral into the sum of two divergent integrals? You seem to split it into the sum of a divergent and convergent integral. – robjohn Jun 05 '15 at 17:48
  • It is correct to show that an integral is divergent proving that it is the sum of a divergent integral and a convergent one. But I'd not write something like $\infty-\pi/2$ in an exam :) – ajotatxe Jun 05 '15 at 18:02