5
$\begingroup$

Find the value:

$$I=\lim_{t\to0^{+}}\lim_{x\to+\infty}\dfrac{\displaystyle\int_{0}^{\sqrt{t}}dx\int_{x^2}^{t}\sin{y^2}dy}{\left[\left(\dfrac{2}{\pi}\arctan{\dfrac{x}{t^2}}\right)^x-1\right]\arctan{t^{\frac{3}{2}}}}$$

I spent some hours doing it, but have failed.

Thank you!

$\endgroup$
  • 3
    $\begingroup$ 'nice limit' LOL $\endgroup$ – Lost1 Sep 11 '13 at 10:07
  • 1
    $\begingroup$ Just in case: are there is $\arctan (t^\frac{3}{2})$ or $(\arctan t)^\frac{3}{2}$? $\endgroup$ – m0nhawk Sep 11 '13 at 10:18
  • $\begingroup$ Start by finding $\lim_{x\to\infty}\left(\frac{2}{\pi}\arctan\frac{x}{t^2}\right)^x$ $\endgroup$ – Jonathan Y. Sep 11 '13 at 10:18
  • $\begingroup$ @m0nhawk is $$\arctan{(t^{\frac{3}{2}})}$$ $\endgroup$ – user94270 Sep 11 '13 at 10:23
  • 1
    $\begingroup$ "Nice limit"...that sounded like old times' stories about the father inviting the unaware man to enter his house telling him "please do come in and meet my beautiful daughter"...and after the poor man met the girl he had to be hospitalized because the impression and horror hurt his weak heart... $\endgroup$ – DonAntonio Sep 11 '13 at 10:26
3
$\begingroup$

One observes $$lim_{x\to\infty}\left(\frac{2}{\pi}\arctan\frac{x}{t^2}\right)^x = e^{lim_{x\to\infty}x\left(\ln\arctan\frac{x}{t^2}-\ln\frac{\pi}{2}\right)}.$$ In computing that limit I find: $$lim_{x\to\infty}x\left(\ln\arctan\frac{x}{t^2}-\ln\frac{\pi}{2}\right) = lim_{x\to\infty}\frac{\ln\arctan\frac{x}{t^2}-\ln\frac{\pi}{2}}{\frac{1}{x}} =\\ = lim_{x\to\infty}\frac{\frac{1}{\arctan\frac{x}{t^2}}\frac{1}{1+\frac{x^2}{t^4}}\frac{1}{t^2}}{-\frac{1}{x^2}} = -lim_{x\to\infty}\frac{x^2t^2}{\arctan\frac{x}{t^2}(t^4+x^2)} = -\frac{2t^2}{\pi}$$

We can therefore find the original limit as $$\lim_{t\to 0^+}\frac{\sqrt{t}\int_\infty^t\sin(y^2)dy}{\left(e^{-\frac{2t^2}{\pi}}-1\right)\arctan(t^{3/2})} =\\ = \lim_{t\to 0^+}\frac{\frac{1}{2\sqrt{t}}\int_\infty^t\sin(y^2)dy + \sqrt{t}\sin(t^2)}{-\frac{4t}{\pi}e^{-\frac{2t^2}{\pi}}\arctan(t^{3/2}) + \left(e^{-\frac{2t^2}{\pi}}-1\right)\frac{1}{1+t^3}\frac{3}{2}\sqrt{t}} =\\ = \lim_{t\to 0^+}\frac{\frac{1}{2}\int_\infty^t\sin(y^2)dy + t\sin(t^2)}{-\frac{4t^{3/2}}{\pi}e^{-\frac{2t^2}{\pi}}\arctan(t^{3/2}) + \left(e^{-\frac{2t^2}{\pi}}-1\right)\frac{1}{1+t^3}\frac{3}{2}t}$$ and every term but the first one in the numerator tends to zero, so unless I've made some mistake (at this point it's more than likely--so let me know) the limit diverges.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I think $e^x-1\approx x,\arctan{x}\approx x$ $\endgroup$ – user94270 Sep 11 '13 at 11:04
  • $\begingroup$ @nanchangjian, I don't follow. Where's the error? $\endgroup$ – Jonathan Y. Sep 11 '13 at 11:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy