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I am having some trouble with this (undergrad) problem. The Twitter account I found this from was deleted so I unfortunately have not found an answer.

I have tried decomposing into Riemann sum and take the exponential representation of sine and it got too messy. I also see that the polynomial term on the denominator can be bounded with sin(x), but I haven't got it to work without the bound not being tight enough. Can anyone point me in the right direction?

The problem is:

Let $I(n)$ be defined as:

$ I(n) = \int_0^{\frac{\pi}{2}} \frac{e^x}{\left(\left(\frac{2}{\pi}\right) x\right)^n + 1} \cdot \left| \sin(x) \sin(nx) \sin(2nx) \right| \, dx $

where $ 3.14 < \pi < 3.15 $ and $ 2.71 < e < 2.72 $

Show that:

$ \frac{6}{5} < \lim_{n \to \infty} I(n) < \frac{5}{4} \ $

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The limit is $$\lim_{n\rightarrow\infty} \int_0^{\pi/2} \frac{e^x}{\left(2x/\pi\right)^n + 1} \left| \sin(x) \sin(nx) \sin(2nx) \right| \, dx=\frac{2 \left(1+e^{\pi /2}\right)}{3 \pi }=1.23302\cdots.$$

For large $n$ and $0<x<\pi/2$ the factor $|\sin(nx)\sin(2nx)|$ may be replaced by its average $\frac{4}{3\pi}$, while the denominator $\left(2x/\pi\right)^n + 1$ may be replaced by unity, resulting in $$\lim_{n\rightarrow\infty}I(n)=\frac{4}{3\pi}\int_0^{\pi/2} e^x \sin x \, dx=\frac{2 \left(1+e^{\pi /2}\right)}{3 \pi }.$$
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  • $\begingroup$ Wow, that is very short. However, I don't understand / haven't learned why we can replace the non-convergent term with the average in this particular case. Can you elaborate a bit on that or is there a theorem I can refer to? (Side note: If I remember correctly, the bounds on pi and e is because this was a no calculator exam. Evaluating that e^(pi/2) to the third decimal would be a nightmare!) $\endgroup$
    – Eftew
    Commented Oct 24 at 15:29
  • $\begingroup$ none of this is rigorous, but it's what we usually do if the integrand is the product of a rapidly oscillating function $f$ times a slowly varying function $g$; you average $f$ over the oscillations, keeping $g$ constant (since it varies much more slowly than the oscillation period); the error becomes smaller and smaller as the oscillation period is reduced, in this case as $n$ becomes larger and larger. $\endgroup$
    – Carlo Beenakker
    Commented Oct 24 at 15:36
  • $\begingroup$ Oh I see haha. How would I go about solving this in an exam though? Do I split the integral into infinitesimal ranges equivalent to each oscillation and argue about the average so that it results in this? $\endgroup$
    – Eftew
    Commented Oct 24 at 15:46

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