Integral and Inequality

Suppose $f(x)$ is a continuously differentiable function on $[a,b]$ satisfying: $$f(a) = f(b) = 0$$ $$\int_a^b (f(x))^2 dx = 1$$ Then show that: $$\int_a^b x^2 (f'(x))^2 dx \geq \frac{1}{4}$$

Solution

By Cauchy: $$\int_a^b x^2(f'(x))^2 dx = \int_a^b (f(x))^2 dx . \int_a^b x^2(f'(x))^2 dx \geq (\int_a^b xf'(x)f(x)dx)^2$$ The last integral can be evaluated using integration by parts, using $u=x$ and $dv = f'(x)f(x)dx$ which yields $v = (f(x))^2/2$, so that: $$\int_a^b xf'(x)f(x)dx = \frac{bf^2(b) - af^2(a)}{2} - \frac{1}{2} \int_a^b (f(x))^2 dx = -\frac{1}{2}$$ so: $$\int_a^b x^2 (f'(x))^2 dx \geq \frac{1}{4}$$

Having fun with infinite series

1. Warm-up problem: show that
$$1 + \frac{1}{2} + \frac{1}{3} + \cdots = \infty$$

2. Suppose $a_1, a_2, \cdots$ is a sequence of positive numbers such that
$$a_1 + a_2 + \cdots + a_n \leq n^2$$
for all $n$, show that
$$\frac{1}{a_1} + \frac{1}{a_2} + \cdots = \infty$$

3. Suppose $a_1, a_2, \cdots$ is a sequence of positive numbers such that
$$a_1 + a_2 + \cdots + a_n \leq n^2 \log n$$
for all $n$, show that
$$\frac{1}{a_1} + \frac{1}{a_2} + \cdots = \infty$$

Solution

Problem 1

This is a standard textbook proof of the divergence of harmonic series, but the point here is to prepare the reader for the subsequent proofs $$\frac{1}{3} + \frac{1}{4} > \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$ $$\frac{1}{5} + \dots + \frac{1}{8} > \frac{1}{8} + \dots + \frac{1}{8} = \frac{1}{2}$$ and so on. So the original series clearly diverges to infinity. The main crux of the proof here is this assertion: $$\frac{1}{2^n+1} + \dots + \frac{1}{2^{n+1}} > \frac{1}{2^{n+1}} + \dots + \frac{1}{2^{n+1}} = \frac{1}{2}$$ for each $n$.

Problem 2

Similar to the proof above, for each $n$ we have: $$a_{2^n+1} + \dots + a_{2^{n+1}} < 4^{n+1}$$ So by AM-HM we have: $$\frac{1}{a_{2^n+1}} + \dots + \frac{1}{a_{2^{n+1}}} > \frac{4^n}{a_{2^n+1} + \dots + a_{2^{n+1}}} > \frac{1}{4}$$ So the original series is greater than $1/4 + 1/4 + \dots = \infty$

Problem 3

Similar to the proof above, for each $n$ we have: $$a_{2^n+1} + \dots + a_{2^{n+1}} < 4^{n+1} \log (2^n) = n . 4^{n+1}.\log 2$$ So by AM-HM we have: $$\frac{1}{a_{2^n+1}} + \dots + \frac{1}{a_{2^{n+1}}} > \frac{4^n}{a_{2^n+1} + \dots + a_{2^{n+1}}} > \frac{1}{4 \log 2 n}$$ So the original series is greater than $\frac{1}{4 \log2} (1 + \frac{1}{2} + \frac{1}{3} + \dots)$ which is also divergent.

Integral Inequality

Prove that:

$$\left( \int_\pi^\infty\frac{\cos x}{x}\ dx\right)^{2} < \frac{1}{{\pi}^{2}} $$

Solution

Integrate by parts:

$$\int \frac{\cos x}{x} dx = \frac{\sin x}{x} + \int \frac{\sin x}{x^2} dx$$

So
$$\int_\pi^\infty\frac{\cos x}{x}\ dx = \int_\pi^\infty \frac{\sin x}{x^2} dx$$

And
$$| \int_\pi^\infty\frac{\cos x}{x}\ dx| = |\int_\pi^\infty \frac{\sin x}{x^2} dx|$$
$$\leq \int_\pi^\infty \frac{| \sin x |}{x^2} dx$$
$$\leq \int_\pi^\infty \frac{1}{x^2} dx $$
$$= \frac{1}{\pi}$$

Integral

Evaluate $ \int e^x \sec x \tan^2x dx$