∫Ln(x)dx

and

∫xⁿln(x)dx

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We can integrate this expression quite simply by parts. We take f=ln(x) and dg=dx, so that we have df=dx/x and g=x. Thus we obtain

(Eq'n 1)

If that seemed a little too simple (as it did for me), then let's try another example; let's integrate xⁿdx by this method. We have f=xⁿ and dg=dx, so we have df=nx^n-1 and g=x. Thus we obtain

(Eq'n 2)

which we solve easily for

(Eq'n 3)

Now we can take the next step and integrate xⁿln(x)dx. Let f=lin(x) and dg=xⁿdx, so we have g=xⁿ⁺¹/(n+1) and df=dx/x. We thus obtain the integration by parts,

(Eq'n 4)

Example Application:

Stirling's Formula

All too often in statistical thermodynamics we encounter the formula N!, in which N represents a very large number. Fortunately in the actual calculations we need only find lnN!, so we want some way of producing a reasonably accurate approximation of lnN! in terms of some easily calculated function. We have

(Eq'n 5)

the product of multiplying together the first N natural numbers. We have in consequence

(Eq'n 6)

In the sequence of natural numbers Δn=1, so we can rewrite that equation as a Riemannian sum;

(Eq'n 7)

If N takes very large values, we can make the approximation and thereby convert the sum into an integral,

(Eq'n 8)

which compares favorably with the actual Stirling's formula,

(Eq'n 9)

which we obtain from the integral

(Eq'n 10)

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