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

I approached it as

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

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

\(\large{= \infty-\frac{\pi}{2}}\)

\(= \infty\)

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## Comments

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TopNewestFor the original question:\[\] The reasoning is almost correct. The only drawback is using \(\infty\) as a real number.\[\] Assuming \[\int_0^\infty\frac{\cos^2(x)}{x^2}\,\mathrm{d}x\] 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. However, we know that \(\int_0^\infty \dfrac{1}{x^2}dx\) diverges hence the original integral is divergent.

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Is this fine? @Azhaghu Roopesh M

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I guess so . The integral clearly diverges . Plus if one was bored and has some time at hand ,then he can even do an IBP and apply the limits to get the answer .

I can't calculate the integral right now , since I don't have anything to write upon (I'm at a party :P) , but I'll check it once I reach home .

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I'm fascinated by this problem, so here is an alternative method before I retire to bed.

A small explanation:

First Inequality: Since \(cos^2(x)/(x^2)\) is positive, the integral is greater than the integral of the same function over a smaller interval.

Second Inequality: for \(x∈[0,π/4], cos^2(x)≥1/2\).

Third Inequality: the integral of \(x^{−n}\) over \([0,k]\) diverges for \(n≥1\) and \(k>0\).

Hope this helps for now.

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@Ishan Dasgupta Samarendra @Rubayet Tusher @Pi Han Goh

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Sorry, I've to go to bed right now (have FIITJEE tomorrow from 10:30 - 5:30). Will surely respond in detail when I get back tomorrow,\[\]

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ok , thats the point, any way then how to deal with it, i'll wait for tomorrow

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Tanishq, your method and answer should be correct. I am saying so, because Wolfram Alpha is also showing the same result. I also tried it using Integration By Parts but got stuck in the midway.

You can also check out these links, though these are of (sinx)^2/(x^2). Maybe of little use to you ...... As, you've already shown that in your post.

http://math.stackexchange.com/questions/13344/proof-of-int-0-infty-left-frac-sin-xx-right2-mathrm-dx-frac-pi2 http://math.stackexchange.com/questions/141695/how-to-calculate-the-integral-of-sin2x-x2 http://math.stackexchange.com/questions/13344/proof-of-int-0-infty-left-frac-sin-xx-right2-mathrm-dx-frac-pi2?lq=1

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@Rubayet Tusher While Wolfram is indeed a very useful tool, you should not base your results on what it shows. On occasion, I have seen Wolfram mark convergent integrals as divergent. It often marks double limits as divergent too (in fact. in one of his problems, @Brian Charlesworth Sir has mentioned this). \[\] Brian Sir, I am truly sorry for having @mentioned you. I was typing your name and had unknowingly pressed the 'enter' key, leading to a Notification being sent to you. I'm truly sorry for the inconvenience caused, Sir.

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Ishan Dasgupta Samarendra I'm extremely sorry for my that comment. Infact, I was not aware of that fact about Wolfram Alpha. Thanks for informing that. And, is it possible to prove using Integration By Parts? I tried it but couldn't go to the end. If it is possible using IBP, please post it when you get time. Thanks again for pointing out my fault.

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@Rubayet Tusher Hello Rubayet. I'm really sorry for the delay. The last few days have been quite hectic and I had quite forgotten about this. My apologies once again to you.\[\] See, this probably cannot be done using IBP and I'll try to explain why. First here is one general rule which often comes useful in general cases:

Now, assuming we try to Integrate by Parts, we would get a first term which we would have to evaluate and a second term (namely the Integral). Thus, if the original Integral was divergent (and in this case it is), then at least one of the terms must be equal to \(\infty\). However, this would again result in \(\infty\) being used as a real number (the way Tanishq had done), which is incorrect. Consequently, we cannot use Integration by parts for this Integral.

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@Ishan Dasgupta Samarendra No need to apologize or sorry. You should be busy with your JEE Preparation at present. So, no problem for the delay. Thanks for posting it and informing me about this fact. Thanks again.

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@Rubayet Tusher Note that if we have \[\int_a^b f(x)dx\] such that the integral diverges at either \(a\) or \(b\), then we can express the limit at which \(\int_a^b f(x)dx\) diverges as \(t\). For our example, let us assume that \(\int_a^b f(x)dx\) diverges at \(b\). Assuming we can express \(\int_a^t f(x)dx\) in a closed form (after performing whatever technique of Integration you use) which does not contain another Integral, then we can take the limit of \(t\) to \(b\) from the left (since for the entire interval of Integration except at \(b\) the integral is convergent) and if the limit exists and is finite, then the Integral is convergent. \[\] This can be extended to cases where \(a\) or both the limits of the interval of Integration is a point where the integral diverges.

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