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For positive integer $n$, an integral of the form $\int_0^\infty x^n e^{-ax}\,\mathrm{d}x$ will reduce to $\frac{n!}{a^{n+1}}$. This can be shown by successively applying integration by parts and observing that the $\left.f(x)g(x)\right|_0^\infty$ terms vanish. The first such term $f(x)g(x)$ is $\left.-\frac{1}{a}e^{-ax}x^n\right|_0^\infty$. It is clear that an exponential will "beat" any power function as $x$ tends to infinity, but how does one show this analytically given the indeterminate form of the expression in the infinite limit? I presume by application of L'Hôpital's rule $n$ times, but I wanted to verify.

After applying integration by parts $n$ times, the only term that does not vanish is of the form $\frac{n!}{a^n} \int_0^\infty e^{-ax}\,\mathrm{d}x$, which quite simply evaluates to $\frac{n!}{a^n} \left(\left.\frac{-1}{a} e^{-ax}\right)\right|_0^\infty = \frac{n!}{a^{n+1}}$.

However, for non-integer $n$, I am getting stuck after integrating by parts $\lfloor n\rfloor$ times. After consulting tables of integration, I realize that the solution is $\frac{\Gamma(n+1)}{a^{n+1}}$, which makes sense given that factorials for non-integer $n$ are evaluated by the gamma function. However, I am not sure how to show this analytically. If one proceeds as before by successively applying integration by parts ($\lfloor n \rfloor$ times) to reduce the integral $\int_0^\infty x^n e^{-ax}\,\mathrm{d}x$, then the $\lfloor n \rfloor$ terms of the form $\left.f(x)g(x)\right|_0^\infty$ vanish and one is left with a final term of the form $\frac{n!}{a^{n}} \int_0^\infty e^{-ax}x^{n-\lfloor n \rfloor}\,\mathrm{d}x$, where $0<n-\lfloor n \rfloor<1$. This is analogous to the previous case of positive $n\in\mathbb{Z}$, except for the residual $x^{n-\lfloor n \rfloor}$ term. Another application of integration by parts would not help, as it would simply reduce the exponent of $x$ to a negative value, and integration by parts could continue ad infinitum. Rather than continuing integration by parts ad infinitum, the $\Gamma$ function appears. How does one resolve this paradox and thereby arrive at the solution $\frac{\Gamma(n+1)}{a^{n+1}}$? Thank you very much.

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2 Answers 2

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You can use substitution, and see the $\Gamma$ function appears right away. Integration by parts kind of mixes things up since we define factorials for integers by the iterated multiplication, so with any non-integer number you'll have problems, so is better to use $\Gamma$ directly:

$$ax = u$$

$$\frac{1}{{{a^{n + 1}}}}\int\limits_0^\infty {{u^n}{e^{ - u}}dx} = \frac{{\Gamma \left( {n + 1} \right)}}{{{a^{n + 1}}}}$$

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Thank you. I think that I didn't realize that the expression was in fact the definition of the gamma function (adjusted by a constant). I thought I had to reduce the power of $x$ (by integration by parts, for instance) to be a real number between $0$ and $1$ in order to invoke the gamma function. –  user001 Feb 17 '12 at 0:30
    
@user001 That's the magic of maths. You can give yourself a name or nickname so people know you. –  Pedro Tamaroff Feb 17 '12 at 0:35
    
So, does the gamma function count as being an analytic solution to the integral posed in the question? If that is the case, I could assign a name to any integral and declare that name to be the analytic solution. Also, why is this particular integral so special to have received a name? (I presume because it is "useful" or "interesting," but maybe there is a more satisfying reason.) Thanks again. –  user001 Feb 17 '12 at 0:54
    
@user001 You should do some research on the Gamma function or ask here, you'll find out it pops up everywhere (I might be exaggerating, but well...) –  Pedro Tamaroff Feb 17 '12 at 1:19

For your first question: Yes, for $a>0$, write $x^n e^{-ax}$ as $x^n\over e^{ax}$ and note that applying L'Hopital's rule $n$-times is justified and will leave you with finding $\lim\limits_{x\rightarrow\infty} {n!\over a^n e^{ax}}$ . Then you can evaluate this limit by other reasoning see that it is 0. (note also when "plugging" in the lower limit $x=0$, you get 0).

So

$$ {-1\over a} x^ne^{-ax} \Bigl|_0^\infty = {-1\over a} \lim\limits_{x\rightarrow\infty} { x^n\over e^{ ax}}-0 = {-1\over a} \lim\limits_{x\rightarrow\infty} {n!\over a^n e^{ax}} =0. $$

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Thank you for confirming my suspicions. –  user001 Feb 17 '12 at 0:27

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