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

This is how a friend suggested I approached it as

$$\large{= \displaystyle \int^{\infty}_{0}\frac{1- \sin^{2} x}{x^2}.dx}$$

$$\large{= \infty-\displaystyle \int^{\infty}_{0}\frac{\sin^{2} x}{x^2}.dx}$$

$$\large{= \infty-\frac{\pi}{2}}$$

$$= \infty$$

Now how he got this result is something I couldn't understand. I would be truly grateful if somebody could please tell me how to integrate this Integral? Many thanks in advance!

  • 1
    $\begingroup$ How did you get that $\int_0^\infty\sin^2x/x^2\,dx=\pi/2$? $\endgroup$ – Tim Raczkowski Jun 5 '15 at 17:32
  • $\begingroup$ What's the point? $\endgroup$ – Ron Gordon Jun 5 '15 at 17:32
  • $\begingroup$ The reasoning looks correct. $\endgroup$ – Akiva Weinberger Jun 5 '15 at 17:35
  • $\begingroup$ @TimRaczkowski Sir, I was editing it. Please could you help me, Sir? $\endgroup$ – Ishan Jun 5 '15 at 17:36
  • $\begingroup$ @RonGordon Sir, I just editted the problem. Please could you help? $\endgroup$ – Ishan Jun 5 '15 at 17:36

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.

  • $\begingroup$ Thanks very, very much for the explanation Sir! Just one last doubt, Sir. How can we split an integral into 1(or more) divergent integrals? Is this allowed? If so, could we split $$\displaystyle\int_{0}^{\infty} \dfrac{1}{x} \left(\tan^{-1}(\pi x) - \tan^{-1}x\right)dx.$$ into $$=\displaystyle\int_{0}^{\infty} \dfrac{tan^-1 (\pi x)}{x} dx -\int_0^\infty \dfrac{\tan^{-1}x}{x}dx.$$ Why, why not? Please help, Sir as I'm quite confused. Many thanks! $\endgroup$ – Ishan Jun 5 '15 at 18:04
  • $\begingroup$ 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. $\endgroup$ – robjohn Jun 5 '15 at 18:30
  • $\begingroup$ 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 $[a,\pi a]\to0$. $\endgroup$ – robjohn Jun 5 '15 at 18:42
  • $\begingroup$ Sir, actually, the integral $\displaystyle\int_{0}^{\infty} \dfrac{1}{x} \left(\tan^{-1}(\pi x) - \tan^{-1}x\right)dx$ was from here and was to be evaluated using Differentiation Under the Integral Sign. In her answer, DR MV had said that we $cannot$ split a convergent integral into 2 divergent integrals. But from your answer, it seems we can. Please could you clarify this doubt of mine, Sir? $\endgroup$ – Ishan Jun 5 '15 at 19:16
  • $\begingroup$ Sir, I was unable to understand what you did to get to the second and third steps in the RHS (particularly what you did to the limits of the interval of integration and how that affected the integrand). Could you please explain these steps to me? Many thanks Sir! $\endgroup$ – Ishan Jun 5 '15 at 19:23

This looks a little better, I'd say.


  • $\begingroup$ Sir could you explain the steps which you took to get to this result please? EDIT: Sir, I still couldn't get it.I'm sorry Sir. $\endgroup$ – Ishan Jun 5 '15 at 17:42
  • $\begingroup$ 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 $k>0$. $\endgroup$ – ajotatxe Jun 5 '15 at 17:44
  • $\begingroup$ Alright Sir. Thanks for your answer. But could you tell me what was wrong, or how the proposed solution is correct? Also, how can we split an integral into 1 (or more) divergent Integrals? $\endgroup$ – Ishan Jun 5 '15 at 17:47
  • $\begingroup$ @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. $\endgroup$ – robjohn Jun 5 '15 at 17:48
  • $\begingroup$ @robjohn Sir that's why I wrote '1(or more)'. $$$$True Sir, I split it into a divergent and a convergent Integral. But is this allowed Sir? $\endgroup$ – Ishan Jun 5 '15 at 17:49

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