Problem 4

Exercise 10.4 (DCT in two ways)   Evaluate the limit

$$\lim_{n\to\infty} \int_0^1 \frac{ndx}{(1+nx)^2(1+x+x^2)}$$ (10.19)

Proof. Substitute $u=nx$ in the integral

$$\int_0^1 \frac{ndx}{(1+nx)^2(1+x+x^2)} = \int_0^n \frac{du}{(1+u)^2(1+u/n+(u/n)^2)}$$ (10.20)

$$= \int_{\mathbb{R}} \frac{1_{[0,n]}(u)du}{(1+u)^2(1+u/n+(u/n)^2)}$$ (10.21)

We have the inequality

$$u\in [0,n] \implies 1/(1+u/n+(u/n)^2) <1$$ (10.22)

Therefore, the integrand is bounded by $1/(1+u)^2\cdot 1_{[0,\infty)}$, which is integrable, so we may apply DCT.

$$\lim_{n\to\infty} \int_{\mathbb{R}} \frac{1_{[0,n]}(u)du}{(1+u)^2(1+u/n+(u/n)^2)}= \int_0^\infty \frac{1}{(1+u)^2}du=1$$ (10.23)

This could also have been proven with integration by parts as follows. Separate with some parentheses

$$I_n = \int_0^1 \left(\frac{n}{(1+nx)^2}\right)\frac{1}{1+x+x^2}dx$$ (10.24)

to reveal that the integrand can be rewritten for an integration by parts as follows:

$$I_n = \int_0^1 \left(-\frac{1}{1+nx}\right)'\frac{1}{1+x+x^2}dx$$ (10.25)

$$= \left.\left(-\frac{1}{1+nx}\right)\frac{1}{1+x+x^2} \right\vert... ...1 -\int_0^1 \left(-\frac{1}{1+nx}\right) \left(-\frac{1+2x}{(1+x+x^2)^2}\right)$$ (10.26)

$$= \left(-\frac{1}{1+n}\right)\frac{1}{1+1+1^2} + 1 - \int_0^1 \frac{1+2x}{(1+nx)(1+x+x^2)^2}$$ (10.27)

The integrand decays in $n$ which is not tied to $1/x^2$, so that limit/integral interchange applies, proving that

$$\lim_{n\to\infty} I_n = 1$$ (10.28)

If one desires a concrete dominating function to supply to the DCT, they can take 2. $\qedsymbol$