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I love doing mathématique which are a little bit hard for my level to challenge myselft and push my mathématiques boundaries. And I come across a problem I can't understand how to finish.

I have to prove that :

$$\left |f(x) - \frac{1}{2x} \right | = \int_{x}^{2x} \frac{t^2 + 1}{t^2\sqrt{t^4+t^2+1}(t^2 + \sqrt{t^4+t^2+1})}dt$$ with, $$f(x) = \int_{x}^{2x} \frac{1}{\sqrt{t^4+t^2+1}}dt$$

The answer is : $$\left |f(x) - \frac{1}{2x} \right | = \left | \int_{x}^{2x} \frac{1}{\sqrt{t^4+t^2+1}}dt - \int_{x}^{2x} \frac{1}{t^2}dt \right |$$ To obtaint the final form, you have to develop and do some integration calculus, not very hard.
I don't understand how to find that : $$\frac{1}{2x} = \int_{x}^{2x} \frac{1}{t^2}dt$$

pdubs
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  • Il fallait un peu d'inspiration. Mais de façon générale, ce qu'on voit ici, c'est qu'on essaye de démontrer que la différence entre $f(x)$ et $1/2x$ est petite (avant tout pour les grandes valeurs de $x$). C'est-à-dire qu'on essaye d'approcher l'intégrale. L'intégrale elle-même semble compliquée à calculer, alors on remplace l'intégrande par une fonction simple, $1/t^2$, qu'on sait intégrer et qui est très proche de l'intégrande pour les grandes valeurs de $t$. Comment peut-on y penser? En faisant $(t^4 + t^2 + 1)^{-1/2} \approx (t^4)^{-1/2} = 1/t^2$. – user49640 May 31 '17 at 16:47
  • C'est une façon très ingénieuse de répondre au problème, merci de votre réponse ! – pdubs May 31 '17 at 16:56
  • If I may ask why would you want to find $\frac{1}{2x}=\int_x^{2x}\frac{1}{t^2}$ it seems easier to go in other direction i.e to simplify the integral and then get the final form instead of writing $\frac{1}{2x}=\int_x^{2x}\frac{1}{t^2}$ – kingW3 May 31 '17 at 17:16
  • @kingW3 In the problem statement, pdubs was only given the first two equations. – user49640 May 31 '17 at 17:22
  • @user49640 Yeah I get that, I still feel that that the right integral (from the first equation) is easier to manipulate to get the third equation( "The answer is:") then to manipulate left side of the first equation to look like the right side (though I may be mistaken). Anyway just wanted to point out to OP that I feel that it's easier to get from right side to left then from left to the right. – kingW3 May 31 '17 at 17:33
  • @kingW3 Yes, you're right. I see what you mean now. I guess you just need to decide whether the expression of the form $|A|$ should be regarded as being $-A$ or $A$. This would be a good answer to the question. – user49640 May 31 '17 at 17:49

2 Answers2

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Write $\dfrac1{t^2}$ as $t^{-2}$ and apply the power rule for integrals.

$$\int_x^{2x} \dfrac1{t^2} \, dt = \int_x^{2x} t^{-2} \, dt = -t^{-1} \bigg|_x^{2x} = -\frac1t\bigg|_x^{2x} = -\left(\frac1{2x}- \frac1x\right) = \frac1{2x} $$

  • Thank you for your quick answer, but I think I misspoke, I don't understand how to find $\int_{x}^{2x} \frac{1}{t^2}dt$ when I only have $\frac{1}{2x}$. – pdubs May 31 '17 at 16:26
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This is simply the integration of a polynomial $t^{-2}$, so just use the rule of "add 1 to the power, divide by the new power" to get the integral to be $-t^{-1}$. $$\int\frac1{t^2}\,dt=-\frac1t+c$$ Applying limits of $2x$, $x$: $$\left[c-\frac1t\right]_x^{2x}=\left(c-\frac1{2x}\right)-\left(c-\frac1x\right)=\frac1x-\frac1{2x}=\frac{2-1}{2x}=\frac1{2x}$$

EDIT:

Oh I see, so you want to jump from having $\frac{1}{2x}$ back to an integral? This would probably need a bit of guess work - you see that $f(x)$ has an integral with limits $2x$ and $x$, so you can set this up:

$$\frac{1}{2x}=\int_x^{2x}g(t)\,dt$$ for some $g(t)$. Then suppose $g(t)$ has antiderivative $G(t)$, so we get $$\frac1{2x}=\int_x^{2x}G'(t)\,dt=\left[G(t)\right]_x^{2x}=G(2x)-G(x)$$ Then solving this (I am not sure how to do so systematically) would give you the solution you're after. I'd say to try a polynomial solution since $\frac1{2x}$ is a power of $x$. So try $G(x)=Ax^\alpha\implies 2^\alpha Ax^\alpha-Ax^\alpha=\frac1{2x}\implies \alpha=-1, \frac A2-A=\frac12\implies A=-1$.

Thus $G(t)=-\frac1t\implies g(t)=\frac 1{t^2}$ works.

John Doe
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