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I am trying to show that:

$\displaystyle\lim_{\lambda \rightarrow \infty} \int_{0}^{\infty} e^{-x}\cos(x)\arctan(\lambda x) \ dx = \dfrac{\pi}{4}$

I've tried using MCT/DCT but haven't found a suitable expression for the above to use either. I'm also somewhat stuck because assuming the integral and limit can be swapped I don't know exactly how to evaluate $\displaystyle\lim_{\lambda \rightarrow \infty} \arctan(\lambda x)$

Thanks

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    $\begingroup$ What is $\lim\limits_{t\to\infty} \arctan t$? $\endgroup$ – Daniel Fischer Mar 21 '14 at 16:07
  • $\begingroup$ @DanielFischer $\dfrac{\pi}{2}$ $\endgroup$ – Noble. Mar 21 '14 at 16:09
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    $\begingroup$ So, for $x > 0$, what is $\lim\limits_{\lambda\to\infty} \arctan (\lambda x)$? $\endgroup$ – Daniel Fischer Mar 21 '14 at 16:09
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    $\begingroup$ @DanielFischer Yeah, it'll be the same $\dfrac{\pi}{2}$. So am I right in thinking I just have to justify why the limit can be bought in, then take the $\dfrac{\pi}{2}$ out of the integral and evaluate $\int_{0}^{\infty} e^{-x}\cos(x) \ dx$? $\endgroup$ – Noble. Mar 21 '14 at 16:14
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    $\begingroup$ Yes. A dominating function to justify it isn't hard to find. $\endgroup$ – Daniel Fischer Mar 21 '14 at 16:15
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We have $$\lim_{\lambda \to\infty} \arctan(\lambda x)=\left\{\begin{array}\\\frac\pi2&\text{if}&x>0\\0&\text{if}&x=0\end{array}\right.\;\;=\frac\pi2\;\text{a.e.}$$ and $$\left|e^{-x}\cos(x)\arctan(\lambda x)\right|\le\frac\pi2 e^{-x}\in L^1([0,\infty))$$ so by the dominated convergence theorem we have $$\displaystyle\lim_{\lambda \rightarrow \infty} \int_{0}^{\infty} e^{-x}\cos(x)\arctan(\lambda x) \ dx =\frac\pi2\int_{0}^{\infty} e^{-x}\cos(x) \ dx=\Re \frac\pi2\int_{0}^{\infty} e^{(-1+i)x} \ dx = \dfrac{\pi}{4}$$

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