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

Alumni Liaison

To all math majors: "Mathematics is a wonderfully rich subject."

Dr. Paul Garrett