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I am learning about integrals with parameters but I'm completely confused by them.

I am confused by how to evaluate this limit

$$ \lim_{R \to \infty} \int_0^{\frac{\pi}{2}} e^{-R \sin{x}} dx $$

I can see that the function $f(x, R) = e^{-R \sin{x}}$ is continuous and also that the limits of integration are continuous. This means that $F(R) = \int_0^{\frac{\pi}{2}} e^{-R \sin{x}} dx$ is also continuous. But I have no idea what to do with that. I would appreciate it a lot if you explain it in details, since I have a very hard time understanding integrals with parameters.

Jesus
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3 Answers3

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On the interval $[0,\pi/2]$, you have $\sin x\geq x/2$. Then $$\tag1 e^{-R\sin x}\leq e^{-Rx/2}. $$ Thus $$ 0\leq \int_0^{\pi/2}e^{-R\sin x}\,dx\leq \int_0^{\pi/2}e^{-Rx/2}\,dx=\frac2R\,\big(1-e^{-R\pi/2}\big)\xrightarrow[R\to\infty]{}0. $$

Martin Argerami
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When you integrate with respect to a variable in a definite bounds, you kill one of the functional dependencies the original function had. Example:

$$ \int_0^1 xt dx = t \int_0^1 x dx =t \frac{x^2}{2}|_1 = \frac{t}{2}$$

Suppose we had a $f(x,t)=xt$ then in the end after integration we get a function with one variable integrated out $F(t) = \frac{t}{2}$ (here being x).

The question is what the value of the integral of this one variable function turns into as you raise $R$ to infinity.

There are two ways to solve the problem, the first would be integrate with $x$ and then take the limit, or you can swap the limit and integral. For the latter, you would need some theorems of analysis to justify the move.

tryst with freedom
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  • Can I swap the integral and limit in my case? Why can I do that? The theorems we've covered talk about when the integral is continuous, integrable and differentiable. I can't figure out why I could move the limit in though. The function that is continuous is F(R), not the integral as a function. – Jesus Jan 11 '22 at 14:10
  • The integral is the F(R) only. The integral is continous with respect to R. – tryst with freedom Jan 11 '22 at 14:29
  • Because $F$ is continuous, $\lim_{R \to \infty} F(R) = F(\lim_{R \to \infty} R)$, which in itself is pretty useless. Are you saying that $F(\lim_{R \to \infty} R) = \int_0^{\pi / 2} \lim_{R \to \infty} f(R, x)$ ? One of the limits isn't even a limit of one variable, how can they be the same? – Jesus Jan 11 '22 at 14:33
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Hint: sometimes is just a matter of estimations. The integral is positive and the function sin is concave in $\left(0,\frac{\pi}{2}\right)$.

Bernkastel
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  • I'm not asking for a trick or a hint, but rather an explanation of how to come to the conclusion that I can swap the limit and the integral. – Jesus Jan 11 '22 at 14:11
  • @Matthew: I don't want to sound rude, but nowhere in your question is mentioned the interchange of the limit with the integral. You've only asked how to evaluate the limit, and I gave you an hint to do that according to the request. Because of this, I've given this hint because it is not always the case that you must interchange limit and integral, but sometimes you can solve your problem with an estimation. – Bernkastel Jan 11 '22 at 14:41
  • Sorry if I wasn't clear enough. Could you help me understand why $F(\lim_{R \to \infty} R) = \int_0^{\pi / 2} \lim_{R \to \infty} f(R, x)$ ? – Jesus Jan 11 '22 at 14:54