POTW 8

Couple ways to go about this.

First, you forgot the constant of integration, which in this case could be solved for to cancel the 1.

Related: any definite integral using this equation will just subtract out the 1.

Not sure if this is significant, but how did the negative sign disappear in the second to last step?

It moved out of the integral.

@Another Matt: Your resolution works for the definite integral case, but for indefinite integrals, I suspect something more subtle is going on.

Consider the case of u = x^a and dv =dx. (Of which the given problem is the special case for a = -1.) You get [x^(a+1)]/(a+1) on the left and [x^(a+1)]/[1 - a/(a+1)] on the right, and those are always equal, except when a = -1. No appeal to arbitrary constants of integration is needed.

What happens when a = -1 is that the result of the integration is ln(x) rather than a constant times x^(a+1). ln(x) has an essential singularity rather than a pole at the origin, and another one at infinity. So I suspect the real issue is that the problem as stated involves being sloppy with adding and subtracting infinities.

The derivation of the integration by parts formula in Wikipedia assumes that u(x) and v(x) are continuously differentable functions. 1/x is not such a function, being discontinuous at x = 0, with different limits from above and below.

This may have something to do with the problem here.

Remember that integration by parts is derived from the product rule in differentiation:

(f(x)*g(x))' = f(x)*g'(x) + f'(x)*g(x)

solve for f(x)*g'(x) and integrate both sides, and you get integration by parts.

Now suppose f(x) = x and g(x) = 1/x. Product rule gives:

(a)' = x*(-1/x^2) + 1*(1/x) = 0

(a)' = 0, so if you solve the equation, you just get 1/x = 1/x.

If you want to be silly, I suppose you could solve it as 1/x = 0 + 1/x, and then integrate both sides:

int(a/x) = C + int(a/x)

but that C just comes out in the wash when you actually do find the antiderivative of int(a/x):

int(a/x) = C + ln(x) + C = ln(x) + C

Since each C in C + ln(x) + C can vary, the result is the same class of functions as ln(x) + C. And this would be trivially true for any function.

This would be the only case where you worry about the constant of integration during integration by parts, nor with limits of integration, since no limits are given.

The problem is that the integral of 1/x is not another rational function, it is defined as the natural logarithm function. Repeated use of the method shown above is sure to fail for that reason.

Looks like my solution to POTW 8 wasn't horrible!

POTW 9 is out now (and it's easier).

Yep, POTW 9 is easy. Consider the case of n = 2. You know that h_1 is a set of horses all of the same color, and h_2 is another set of horses all of the same color, but these two sets do not overlap, so there is no guarantee that h_1 and h_2 are the same color.

POTW 9 is a standard exercise that arises in discrete mathematics classes, where students are being introduced to induction for the first time. It often happens that students will carelessly provide an argument for the inductive step that does not quite cover every case.

If I remember how to calculate odds, the odds against the third horse are 11:4.