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I saw a solution to the problem

$$ \lim_{\alpha \to 0} \int_{\alpha}^{1+\alpha} \frac{dx}{1 + x^2 + \alpha^2}$$

that involved simply by moving the limit inside the integral which simplifies to an elementary integral that further gives the limit of $\frac\pi 4$. That is:

$$\lim_{\alpha \to 0} \int_{\alpha}^{1+\alpha} \frac{dx}{1 + x^2 + \alpha^2} = \int_{\alpha}^{1+\alpha} \lim_{\alpha \to 0} \frac{dx}{1 + x^2 + \alpha^2} = \int_{0}^{1} \frac{dx}{1 + x^2} = \arctan1 - \arctan0 = \dfrac \pi 4 $$

There are no remarks explaining why it is allowed to move the limit in this case. As far as I know, the monotone convergence theorem and the dominated convergence theorem allow us to do this when we have infinite limits, as explained in the answers here. In this case, by what theorem is it allowed?

E.Nole
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    See also: https://math.stackexchange.com/questions/3412510/using-dominated-convergence-theorem-with-continuous-limits, https://math.stackexchange.com/questions/2681712/extending-the-dominated-convergence-theorem-to-continuous-limits. – Jan Mar 02 '20 at 12:38
  • If you want to use the DCT, you better mention a dominating function. – Thorgott Mar 02 '20 at 12:49
  • @Thorgott I know that. My question is that, since the limit in the question is not to infinity, as is in the condition for DCT, doesn't that mean we can't use DCT in this case? – E.Nole Mar 02 '20 at 12:58
  • My remark was just for the sake of completeness. See Calvin's answer for the main question. – Thorgott Mar 02 '20 at 12:59

1 Answers1

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2 things -

  1. you need a change of variables $y=x-\alpha$ to move the $\alpha$ out of the integration limits. In particular $\int_\alpha^{1+\alpha} \lim_{\alpha\to 0}$ is nonsense. An alternative is to write the integration limits as multiplication with an indicator $\mathbb 1_{[\alpha,1+\alpha]}$

  2. To apply the usual sequential DCT/MCT, use the sequential formulation of the limit. Let $\alpha_n \to 0$ be arbitrary. Then what you need is $\lim_{n\to\infty} \int_0^1 \frac{dy}{1+(y+\alpha_n)^2 +\alpha_n^2}$ which is amenable to textbook versions of the theorems

Calvin Khor
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  • What is the sequential formulation of the limit ? – E.Nole Mar 02 '20 at 13:00
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    $f(x)\to c$ as $x\to y$ iff for every sequence $x_n\to y$ as $n\to\infty$, we have $f(x_n) \to c$ as $n\to\infty$ @E.Nole – Calvin Khor Mar 02 '20 at 13:01
  • Could you fill in the steps starting from the change of variable? I'm still not quite clear on how to get to the form in step 2. – E.Nole Mar 02 '20 at 13:10
  • When I set $y=x+\alpha$ I still have $\alpha$ in the limits i.e $\lim_{y \to x} \int_{2\alpha}^{1+ 2 \alpha} \frac{dy}{1+(y-\alpha)^2 +\alpha^2}$ – E.Nole Mar 02 '20 at 13:16
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    @E.Nole that should have been a minus sign not a plus sign, sorry – Calvin Khor Mar 02 '20 at 13:19