4.19
Let $ \alpha \geq 1 $ and compute, with justification,
$ \lim_{n\rightarrow\infty} \int_{0}^{\pi} n\ln\left[1+\left(\frac{\sin{x}}{n}\right)^\alpha\right] dx $
Solution
So, first, we state (without proof) an inequality (Jensen's Inequality).
Let $ f $ be a convex function defined on an interval $ [a,b] $, and g a integrable function on $ [a,b] $. Then:
$ f\left(\int_{a}^{b} g\right) \leq \int_{a}^{b} \left(f \circ g\right) $
Now, note that $ -\ln x $ is convex, so for $ f = -\ln x $, we get
$ -f\left(\int_{a}^{b} g\right) \geq \int_{a}^{b} \left(-f \circ g\right) $
Now, we need to pull the limit inside the integral, so we proceed as follows:
$ \int_{0}^{\pi} n\ln{\left[1 + \left( \frac{\sin{x}}{n}\right)^\alpha \right]} dx = \int_{0}^{\pi} \ln{\left[1+\left(\frac{\sin{x}}{n}\right)^\alpha\right]^{n}} dx \leq \ln\int_{0}^{\pi} \left[1+\left(\frac{\sin{x}}{n}\right)^\alpha\right]^{n} dx $
Now, since $ \alpha \geq 1 $, we have the following:
$ \int_{0}^{\pi} \left[1+\left(\frac{\sin{x}}{n}\right)^\alpha\right]^{n} dx \leq \int_{0}^{\pi} \left[1+\frac{\sin{x}}{n}\right]^{n} dx $
And the integrand on the right converges uniformly to $ \exp{\left(\sin{x}\right)} $, so
$ \int_{0}^{\pi} \left[1+\frac{\sin{x}}{n}\right]^{n} dx \leq \int_{0}^{\pi} e dx = e \pi $
And we choose, as our dominating function, $ \exp{\pi} $
And now, by the Dominating Convergence Theorem, we can pull the limit inside the integral.
So now, we start computing our limit, using L'Hospital's Rule.
2 cases:
$ \alpha = 1 $
Only consider the integrand in computing the limit function.
$ n\ln\left[1+\left(\frac{\sin{x}}{n}\right)^\alpha\right] = \frac{\ln\left[1+\left(\frac{\sin{x}}{n}\right)^\alpha\right]}{1/n} $
Using L'Hospital's Rule, you get the limit simplifying to
$ \frac{n\sin{x}}{n+\sin{x}}\rightarrow \sin{x} $
So the integral is
$ \int_{0}^{\pi} \sin{x} dx = -\cos{x}|_{0}^{\pi} = 2 $
---
$ \alpha > 1 $
Only consider the integrand in computing the limit function.
$ n\ln\left[1+\left(\frac{\sin{x}}{n}\right)^\alpha\right] = \frac{\ln\left[1+\left(\frac{\sin{x}}{n}\right)^\alpha\right]}{1/n} $
Using L'Hospital's Rule, you get the limit simplifying to
$ \frac{n\alpha\sin{x}}{n^\alpha+\sin{x}^\alpha} \rightarrow 0 $
So the integral is clearly 0.
Written by Nicholas Stull
