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Find $$ \lim_{a\to \infty} \frac{1}{a} \int_0^{\infty}\frac{x^2+ax+1}{1+x^4} \arctan\left(\frac{1}{x}\right)dx $$


I tried to find $$ \int_0^{\infty} \frac{x^2+ax+1}{1+x^4}\arctan\left(\frac{1}{x}\right) dx $$ Let $$ I(a) = \int_0^\infty \frac{x^2+ax+1}{1+x^4}\arctan\left(\frac{1}{x}\right) dx $$ Let $$ \begin{split} t &= \arctan(1/x) \\ \frac{dt}{dx} &= \frac{-1}{1+x^2} \\ dt &= \frac{-1}{1+x^2}dx \end{split} $$

I am stuck here, there seems no way to further solving.

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    $\begingroup$ ...Hence the idea to compute $I(a)$ for every $a$ might not have been the most direct one. Any idea to solve the question asked, that is, to compute the limit when $a\to\infty$? For example, what would be the limit of $$\frac{1}{a} \int_0^{\infty}\frac{x^2+1}{1+x^4} \arctan\left(\frac{1}{x}\right)dx\ ? $$ $\endgroup$ – Did Mar 25 '16 at 14:46
  • $\begingroup$ The limit is $$\frac{\pi^2}{16}$$ $\endgroup$ – Turing Mar 25 '16 at 14:50
  • $\begingroup$ Yes,correct.How did you find this@1over137 $\endgroup$ – Brahmagupta Mar 25 '16 at 14:52
  • $\begingroup$ @Did you ask the right questions $\endgroup$ – tired Mar 25 '16 at 16:29
  • $\begingroup$ @tired Thanks (unfortunately the suggestion did not cause much reaction from the OP...). $\endgroup$ – Did Mar 25 '16 at 21:50
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One may observe that, as $a\to \infty$, $$ \begin{align} &\frac1a \int_0^{\infty}\frac{x^2+ax+1}{1+x^4} \arctan\left(1/x\right)\:dx- \color{red}{\int_0^{\infty}\!\!\frac{x}{1+x^4} \arctan\left(1/x\right)\:dx} \\\\&=\frac1a\int_0^{\infty}\frac{x^2+ax+1}{1+x^4} \arctan\left(1/x\right)\:dx- \frac1a\int_0^{\infty}\frac{ax}{1+x^4} \arctan\left(1/x\right)\:dx \\\\&=\frac1a\int_0^{\infty}\frac{x^2+1}{1+x^4} \arctan\left(1/x\right)\:dx \: \longrightarrow \: 0,\tag1 \end{align} $$ since the latter integral is convergent.

On the other hand, we have $$ \int_0^{\infty}\frac{x}{1+x^4} \:\arctan\left(1/x\right)\:dx=\int_0^{\infty}\frac{x}{1+x^4}\: \arctan\left(x\right)\:dx\quad (x \to 1/x) $$ using, $\displaystyle \arctan\left(x\right)+\arctan\left(1/x\right)=\frac{\pi}2$ ($x>0$), one gets $$ \begin{align} &\int_0^{\infty}\!\!\frac{x}{1+x^4}\: \arctan\left(x\right) \:dx+\int_0^{\infty}\!\!\frac{x}{1+x^4}\:\arctan\left(1/x\right)\:dx \\\\& =\frac{\pi}2\int_0^{\infty}\!\!\frac{x\:dx}{1+x^4} \\\\&=\frac{\pi}4\int_0^{\infty}\!\!\frac{du}{1+u^2} \\\\& =\frac{\pi^2}8, \end{align} $$ thus $$\displaystyle \color{red}{\int_0^{\infty}\!\frac{x}{1+x^4}\:\arctan\left(1/x\right)\:dx}=\frac{\pi^2}{16}$$ giving, from $(1)$,

$$ \lim_{a\to \infty} \frac{1}{a} \int_0^{\infty}\frac{x^2+ax+1}{1+x^4}\: \arctan\left(\frac{1}{x}\right)dx=\frac{\pi^2}{16}.\tag2 $$

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  • $\begingroup$ Is it always okay to take the limit of a before integrating? I imagine it is the same to take it before or after, but I'm not sure $\endgroup$ – Fede Poncio Mar 25 '16 at 22:30
  • $\begingroup$ @FedePoncio To avoid considerating limit before or after integration here, I've just proved that $I(a)$ tends to the integral in red (since their difference tends to $0$, see $(1)$). Then I've evaluated the integral in red. Thanks. $\endgroup$ – Olivier Oloa Mar 25 '16 at 23:21
  • $\begingroup$ I have taken the liberty to fix a minor typo. +1 for the nice solution. $\endgroup$ – Paramanand Singh Mar 26 '16 at 3:39
  • $\begingroup$ @ParamanandSingh Thank you! $\endgroup$ – Olivier Oloa Mar 26 '16 at 4:03

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