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I need to find

$$f(n) = \int^\infty_0 t^{n-1} e^{-t} dt$$

So I think I find the indefinate integral first? But what do I do with $n$, since I am integrating with respect to $t$?

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    $e^{-t}$. $ $ $ $2012-04-10
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    Are you certain the exponent on $e$ is $-1$?2012-04-10
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    @AlexBecker, corrected it. It should be $t$. Its supposed to be the gamma function2012-04-10
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    @JiewMeng IIRC, for integers $n$ you use integration by parts.2012-04-10

5 Answers 5

1

I think $n$ is parametr. So you have to do it, treating $n$ like normal number.

6

Here is also another approach, though a posteriori.

Tonelli's theorem enables us to exchange the order of integration and summation of a sequence of nonnegative functions. Thus for $ 0 < r < 1 $, $$\begin{align*} \sum_{n=0}^{\infty} \frac{r^n}{n!} \int_{0}^{\infty} x^n e^{-x} \; dx & = \int_{0}^{\infty} \sum_{n=0}^{\infty} \frac{(rx)^n}{n!} e^{-x} \; dx = \int_{0}^{\infty} e^{-(1 - r)x} \; dx \\ & = \frac{1}{1 - r} = \sum_{n=0}^{\infty} r^{n}. \end{align*}$$

Actually, it's essentially identical to cooper.hat's approach, since it is just the Taylor expansion of $f(1-r)$ with his/her $f$.

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    Very cute. Took me a moment to compare coefficients of $r^n$. The key is the uniqueness of expansion (for those reasoning at my pedestrian pace).2012-04-11
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    @sos440 Great contribution.2012-04-11
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    This is a nicer proof, it computes all coefficients in one swoop.2012-04-11
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    Thanks. What I'm concerned is, howerer, that my solution seems not an adequate answer since we have to know that $\Gamma(n+1)=n!$ prior to the solution. In view of this deficiency, integration-by-parts solution seems to be most illuminating.2012-04-11
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    Very nice proof. Is it your own? Integration by parts is doubtless more illuminating for those who've never seen this proposition proved before, but maybe this one's more illuminating to those who have long known that.2012-04-11
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    Yes, I devised it by myself, though I'm also sure that this must have been known long before.2012-04-11
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This integral can be easily computed by means of integration by parts. Let's call it $$\Gamma(n) =\int\limits_0^\infty t^{n-1}e^{-t}dt$$

We want to show that $\Gamma(n)$ follows some law.

Let's integrate by parts, with $u=t^{n-1}$. Then

$$\int\limits_0^\infty t^{n-1}e^{-t}dt=\left.-e^{-t}t^{n-1}\right|_0^\infty+(n-1)\int\limits_0^\infty t^{n-2}e^{-t}dt$$

but since $\left.-e^{-t}t^{n-1}\right|_0^\infty$ vanishes, we have that

$$\int\limits_0^\infty t^{n-1}e^{-t}dt=(n-1)\int\limits_0^\infty t^{n-2}e^{-t}dt$$ or that

$$\Gamma(n)=(n-1)\Gamma(n-1)$$

We can go on and use the relation to get

$$\Gamma(n)=(n-1)!$$

Another approach is copper's:

Define $$I(x) = \int\limits_0^\infty e^{-tx} dt $$

By means of Leibniz' rule, we have that

$$I^{(n)}(x) = \int\limits_0^\infty (-t)^ne^{-tx} dt $$

So we're interested in

$$(-1)^nI^{(n)}(1) = \int\limits_0^\infty t^n e^{-t} dt $$

But since

$$I(x) = \int\limits_0^\infty e^{-tx} dt=\frac{1}{x}$$ and we can simply show that

$$I^{(n)}(x)=\frac{(-1)^n n!}{x^{n+1}}$$ it follows that

$$(-1)^nI^{(n)}(1)=\frac{ n!}{1^{n+1}}=n!$$ as desired.

4

If $n>0$ is an integer, here is another approach: Let $I(x) = \int^\infty_0 e^{-x t} dt$, with $x>0$. It is straightforward to evaluate $I(x) = \frac{1}{x}$, and notice that $f(1) = I(1) = 1$. To continue, notice that $\frac{d I(x)} {d x} = \int^\infty_0 (-t) e^{-x t} dt$, and by direct computation, $\frac{d I(x)} {d x} = -\frac{1}{x^2}$, so $f(2) = -\frac{d I(1)} {d x} = 1$. The process may be continued by induction.

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    Of course, to be complete, you need to justify the exchange of differentiation and integration.2012-04-11
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$$\int^\infty_0 t^{n-1} e^{-t} dt=\Gamma(n)$$