 Solution

ID: past-exam-of-the-differential-equations-and-applications-course-of-the-university-of-oxford/2010/a/3/iii/solution

Assume first that

n \geq m

. Substituting Rodrigues' formula for

L_{n}

gives

I_{nm} = \frac{1}{n !} \int_{0}^{\infty} L_{m} (x) \frac{d ^{n}}{d x ^{n}} (x^{n} e^{- x}) d x = \frac{( - 1 ) ^{n}}{n !} \int_{0}^{\infty} L_{m}^{(n)} (x) x^{n} e^{- x} d x .

n

e^{- x}

tends to zero; at zero every derivative of

x^{n} e^{- x}

of order less than

n

vanishes. If

n > m

L_{m}^{(n)}

is zero, proving orthogonality. The case

m > n

follows by exchanging the two factors.

For

m = n

, the leading coefficient found above gives

L_{n}^{(n)} = (- 1)^{n}

. Therefore

I_{nn} = \frac{1}{n !} \int_{0}^{\infty} x^{n} e^{- x} d x = \frac{Γ ( n + 1 )}{n !} = 1.

The same conclusion includes

n = 0

directly. Thus the full Laguerre orthonormality relation is

\int_{0}^{\infty} L_{n} (x) L_{m} (x) e^{- x} d x = δ_{nm} .

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