*Originally posted by humy*

**I am frustrated here.
**

I know from experience you always know exactly what you are talking about but I don't understand your notation you are using here. For example, what does;

" [1/p t^p (1 - t)^(q - 1)][x, 0] "

mean? I am confused by the application of those square brackets.

Where did that expression come from?

How does it relate to integration by par ...[text shortened]... tupid here, failed to see anything I know how to use to help actually *solve* the integral.

The square bracket notation is meant to be read as:

[f(x)][a, b] = f(b) - f(a). The usual notation for what I was trying to do puts a and b as sub and superscripts after the first bracket, I was just borrowing your notation from the integral.

I don't recognize your formula for integration by parts. First an identity:

∫[a, b] df/dx dx = ∫[f = f(a), f = f(b)] df = f(b) - f(a) = [f(x)][a, b]

which is the Fundamental Theorem of Calculus. From the product rule for calculus we have:

d(uv)/dx = udv/dx + (du/dx)v = uv' + u'v

So, integrating we have:

∫[a, b] dx (d(uv)/dx) = ∫[a, b] dx (uv' + u'v)

We can use our identity on the left hand side to get:

∫[a, b] dx(d(uv)/dx) = [u(x)v(x)][a, b] = u(b)v(b) - u(a)v(a)

Putting this back into our integral expression and rearranging we get the usual form:

∫[a, b] dx u(x)(dv(x)/dx) = [u(x)v(x)][a, b] - ∫[a, b] dx (du(x)/dx)v(x)

or using the prime notation for derivatives:

∫[a, b] dx u(x)v'(x)= u(b)v(b) - u(a)v(a) - ∫[a, b] dx u'(x)v(x)

The application of this formula is what is normally meant by "Integration by Parts".