# Simplifying an integral by introducing an additional parameter.

This is one of my homework tasks this week.

Calculate the integral $$I = \int_0^\infty dx\ x^3 e^{-x}$$ by introducing an additional parameter $\lambda$ and rewriting the exponential function as $e^{-x} = e^{-\lambda x}$ with $\lambda = 1$. Use the property $de^{-\lambda x} / d\lambda = -xe^{-\lambda x}$ to simplify and calculate the integral.

I calculated the integral using integration by parts as $$\left[-e^{-x}(x^3 + 3x^2 + 6x +6)\right]^\infty_0$$, then using L'Hôpital's rule three times I can calculate $$\lim_{x \to \infty} \frac{x^3 + 3x^2 +6x +6}{e^x} = \lim_{x \to \infty}\frac{6}{e^x} = \frac{6}{e^0} = 6$$

I just can't figure out, why I would need an additional $\lambda$. Can someone give me a hint?

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Let $$I(n; \lambda) = \int_0^{\infty} x^n e^{-\lambda x} dx$$ Then we get that $$I_{\lambda}(n;\lambda) = -\int_0^{\infty}x^{n+1} e^{-\lambda x}dx = - I(n+1; \lambda)$$ Hence, $$I(n+1; \lambda) = - \dfrac{dI(n; \lambda)}{d \lambda} = - \dfrac{d}{d \lambda}\left(-\dfrac{dI(n-1; \lambda)}{d \lambda} \right) = \dfrac{d^2 I(n-1; \lambda)}{d \lambda^2}$$ Proceeding like this (i.e. using induction), we get that $$I(n+1; \lambda) = (-1)^k\dfrac{d^k I(n+1-k; \lambda)}{d \lambda^k} = (-1)^{n+1} \dfrac{d^{n+1} I(0; \lambda)}{d \lambda^{n+1}}$$ Now $$I(0,\lambda) = \int_0^{\infty} e^{-\lambda x} dx = \dfrac1{\lambda}$$ Hence, $$I(n; \lambda) = (-1)^{n} \dfrac{d^n (1/\lambda)}{d \lambda^n} = \dfrac{n!}{\lambda^{n+1}}$$

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You would introduce a parameter as $\lambda$ in this case to glean insight into values of the integral for higher powers of $x$, i.e.
$$\int_0^{\infty} dx \: x^n e^{- \lambda x}$$
which incidentally is $n!/\lambda^{n+1}$.