Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

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, $$I(n; \lambda) = (-1)^{n} \dfrac{d^n (1/\lambda)}{d \lambda^n} = \dfrac{n!}{\lambda^{n+1}}$$

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.