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Integration techniques

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Integration Using Parametric Derivatives

Suppose you wanted to find the integral:

∫ 0 ∞ x 2 e − 3 x d x {\displaystyle \int _{0}^{\infty }x^{2}e^{-3x}dx}

We may solve this by starting with the integral:

∫ 0 ∞ e − t x d x = {\displaystyle \int _{0}^{\infty }e^{-tx}dx=}

[ e − t x − t ] 0 ∞ = {\displaystyle \left_{0}^{\infty }=}

[ lim x → ∞ e − t x − t ] − [ e − t 0 − t ] = {\displaystyle \left-\left=}

[ 0 ] − [ 1 − t ] = 1 t {\displaystyle \left-\left={\frac {1}{t}}}

Now that we know:

∫ 0 ∞ e − t x d x = 1 t {\displaystyle \int _{0}^{\infty }e^{-tx}dx={\frac {1}{t}}}

Suppose we found the second derivative with respect, not to x, but to t:

d 2 d t 2 ∫ 0 ∞ e − t x d x = d 2 d t 2 1 t {\displaystyle {\frac {d^{2}}{dt^{2}}}\int _{0}^{\infty }e^{-tx}dx={\frac {d^{2}}{dt^{2}}}{\frac {1}{t}}}

∫ 0 ∞ d 2 d t 2 e − t x d x = d 2 d t 2 1 t {\displaystyle \int _{0}^{\infty }{\frac {d^{2}}{dt^{2}}}e^{-tx}dx={\frac {d^{2}}{dt^{2}}}{\frac {1}{t}}}

∫ 0 ∞ d d t − x e − t x d x = d d t − 1 t 2 {\displaystyle \int _{0}^{\infty }{\frac {d}{dt}}-xe^{-tx}dx={\frac {d}{dt}}-{\frac {1}{t^{2}}}}

∫ 0 ∞ x 2 e − t x d x = 2 t 3 {\displaystyle \int _{0}^{\infty }x^{2}e^{-tx}dx={\frac {2}{t^{3}}}}

Now notice that this solution takes the same form as the original proposed question. In the original problem, t = 3. Substituting that into our new solution equation:

∫ 0 ∞ x 2 e − 3 x d x = 2 3 3 = 2 27 {\displaystyle \int _{0}^{\infty }x^{2}e^{-3x}dx={\frac {2}{3^{3}}}={\frac {2}{27}}}