Questions 5 (Hints and solutions start on page .)

Q 3.6

Use integration by parts to evaluate the following integrals.

$\displaystyle \int$ x sin(2x) dx, $\displaystyle \int$ x³ln x dx, $\displaystyle \int$ xe^{3x - 1} dx

Now do the following by parts -- where you may have to use parts more than once.

$\displaystyle \int$ x²sin(x) dx, $\displaystyle \int$ x²e^{3x - 1} dx, $\displaystyle \int$ ln²x dx

$\displaystyle \int$ e^xsin x dx, $\displaystyle \int$ arctan(x) dx, $\displaystyle \int$ x²e^-x dx

$\displaystyle \int$ x³e^-x dx, $\displaystyle \int$ x⁴e^-x dx $\displaystyle \int$ x⁵e^-x dx

Q 3.7 This is mainly of interest to Statisticians.

A probability density is a function p(x) that is never negative and satisfies the condition $\displaystyle \int_{-\infty}^{\infty}$ p(x) dx = 1. A random variable X has probability density p(x) if the probability that X takes a value between a and b is given by

Pr(a $\displaystyle \le$ X $\displaystyle \le$ b) = $\displaystyle \int_{a}^{b}$ p(x) dx

where a < b.

The mean value of X is defined to be

$\displaystyle \mu$ = $\displaystyle \int_{-\infty}^{\infty}$ x.p(x) dx

The variance of X is defined to be

$\displaystyle \nu$ = $\displaystyle \int_{-\infty}^{\infty}$ (x - $\displaystyle \mu$ )²p(x) dx

and the standard deviation $\sigma$ is given by $\sigma^{2}_{}$ = $\nu$ .

Here are three functions. Check that they are all probability densities.

$% l2h can't cope with cases \begin{align*} p_1(x) &= \begin{cases}\frac{1}{b-a}... ...) &= \begin{cases}\frac{1}{x^2} & x \ge 1 \\ 0 & x < 1 \end{cases}\end{align*}$

Show that p₁(x) (a uniform distribution) has mean ${\frac{1}{2}}$ (a + b) and variance ${\frac{1}{12}}$ (b - a)².

Show that p₂(x) (a Poisson distribution) has mean 1/ $\theta$ and variance 1/ $\theta^{2}_{}$ .

Show that p₃(x) does not have a finite mean.

Show that the formula for the variance can be rearranged to give

$\displaystyle \nu$ = $\displaystyle \int_{-\infty}^{\infty}$ x²p(x) dx - $\displaystyle \mu^{2}_{}$ .

If the random variable X has the Poisson distribution p₂(x) show that the probability that the value of X is greater than the mean value is 1/e.

*Q 3.8 This is a follow up to some integrals that you did in an earlier exercise. Let

I_n = $\displaystyle \int_{0}^{1}$ xⁿe^x dx

for n a whole number $\ge$ 0. So, for example,

I₆ = $\displaystyle \int_{0}^{1}$ x⁶e^x dx and I₉ = $\displaystyle \int_{0}^{1}$ x⁹e^x dx

By using integration by parts, integrating the exponential and differentiating the power, prove the general result that if n > 0

I_n = e - nI_{n - 1}

This is what is known as a Reduction Formula, it gives us the value of I_n in terms of the value of I_{n - 1}. What use is that? Well, it is easy to evaluate I₀ -- do so. The formula now tells us the value of I₁ and, using it once more, the value of I₂ and so on. By repeated use of the formula, and no further integration, we can get the value of I_n for and positive integer n. Work out I₄.

Go through the same process for the integral $\displaystyle \int_{0}^{1}$ xⁿe^-x dx and work out I₃.

Now look at the integral

I_n = $\displaystyle \int_{0}^{\pi}$ xⁿsin x dx n $\displaystyle \ge$ 0

By doing integration by parts on this twice over, integrating the trig function and differentiating the power, show that

I_n = $\displaystyle \pi^{n}_{}$ - n(n - 1)I_{n - 2} n > 1

This is slightly more complicated in that it relates I_n to I_{n + 2} rather than to I_{n + 1}. This means that you go up in steps of 2. If you think about it you will realise that you now need two starting points: I₀ and I₁. Work out both of these and then work out I₃ and I₄.

*Q 3.9 Consider

I(n, m) = $\displaystyle \int_{0}^{1}$ xⁿ(1 - x)^m dx

where n and m are integers $\ge$ 0. By making the substitution y = 1 - x, show that I(n, m) = I(m, n). Show that

I(n, 0) = I(0, n) = $\displaystyle {\frac{1}{n+1}}$

Use integration by parts to show that

I(n, m) = $\displaystyle {\frac{m}{n+1}}$ I(n + 1, m - 1)

Deduce from this that

I(n, m) = $\displaystyle {\frac{n! m!}{(n+m+1)!}}$