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How can I compute the following limit?

$$\lim_{a \to 0^+} \left(a \int_{1}^{\infty} e^{-ax}\cos \left(\frac{2\pi}{1+x^{2}} \right)\,\mathrm dx\right)$$

Any hints you can please give?

Cheers

  • 0
    I end up with: $\displaystyle \lim_{a\to0+}\int_{a}^{\infty} e^{-y} cos\left( \frac{2\pi}{1+(y/a)^{2}}\right) dy$. Then I'm stuck. How do I apply DCT here?2010-11-06
  • 0
    Well, we have that $\int_{a}^{\infty}=\int_{0}^{\infty}-\int_{0}^{a}=I_1-I_2$. Now $I_2$ goes to $0$ as $a\to 0$. In $I_1$ we should first multiply the numerator and denominator of the fraction under $\cos$ by $a^2$ and then pass to the limit using the DCT.2010-11-06
  • 0
    You're almost there. The integrand is $\le1$, so $\int_0^a$ vanishes. Thus the limit is equal to $\lim\limits_{a\to0^+}\int_0^\infty e^{-y}\cos\left(\frac{2\pi a^2}{a^2+y^2}\right)\;\mathrm{d}y$. See [Didier's answer](http://math.stackexchange.com/a/105963/).2012-02-05

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