# multiple partial integration of one function

While playing around with $$\int _{ 0 }^{ \infty }{ { x }^{ 2 } { e }^{ -2x }dx }$$ I found a very interesting propertie for $$n \in \mathbb{N}$$... $\int _{ 0 }^{ \infty }{ { x }^{ n } { e }^{ -2x }dx } ={ \left( \frac { 1 }{ 2 } \right) }^{ n+1 }\cdot n!$ For me the most amazing fact of this integral is, that the total area (at the non negative x-axis) under the curve of a transcendental exponential curve is rational (for any natural $$n$$).

Here is the proof

$\int _{ 0 }^{ \infty }{ { x }^{ n }\cdot { e }^{ -2x }dx } =\lim _{ c\rightarrow \infty }{ \left( \int _{ 0 }^{ c }{ { x }^{ n }\cdot { e }^{ -2x }dx } \right) }$ Now we can apply partial integration where $u=x^n$ and $v'=e^{-2x}$ : $\int _{ 0 }^{ \infty }{ { x }^{ n } { e }^{ -2x }dx } =\lim _{ c\rightarrow \infty }{ \left( { \left[ { x }^{ n }{ e }^{ -2x } \right] }_{ 0 }^{ c }-\int _{ 0 }^{ c }{ n{ x }^{ n-1 }\cdot \left( -\frac { 1 }{ 2 } \right) { e }^{ -2x }dx } \right) }$ Here we can simplify the integral by "removing" the factors. $\int _{ 0 }^{ \infty }{ { x }^{ n }\cdot { e }^{ -2x }dx } =\lim _{ c\rightarrow \infty }{ \left( { \left[ { x }^{ n }{ e }^{ -2x } \right] }_{ 0 }^{ c }+\frac { 1 }{ 2 } n\int _{ 0 }^{ c }{ { x }^{ n-1 }{ e }^{ -2x }dx } \right) }$

As you can see, we have almost the same integral on the right hand side as on the left hand side of the equation. And now we repeat the whole process: $\int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -2x }dx } =\lim _{ c\rightarrow \infty }{ \left( { \left[ { x }^{ n }{ e }^{ -2x } \right] }_{ 0 }^{ c }+\frac { 1 }{ 2 } n{ \left[ { x }^{ n-1 }{ e }^{ -2x } \right] }_{ 0 }^{ c }-\frac { 1 }{ 2 } n\int _{ 0 }^{ c }{ \left( n-1 \right) { x }^{ n-2 }\cdot \left( -\frac { 1 }{ 2 } \right) { e }^{ -2x }dx } \right) }$

For simplicity lets rename everything besides the integral (and its factor) as "$Z\left( x \right)$". You will see in a minute, why these summands are unimportant... \begin{aligned}\int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -2x }dx } &=\lim _{ c\rightarrow \infty }{ \left( Z\left( x \right) -\frac { 1 }{ 2 } n\int _{ 0 }^{ c }{ \left( n-1 \right) { x }^{ n-2 }\cdot \left( -\frac { 1 }{ 2 } \right) { e }^{ -2x }dx } \right) } \\ &=\lim _{ c\rightarrow \infty }{ \left( Z\left( x \right) +\frac { 1 }{ 4 } n\left( n-1 \right) \int _{ 0 }^{ c }{ { x }^{ n-2 }{ e }^{ -2x }dx } \right) }\\ &=\lim _{ c\rightarrow \infty }{ \left( Z\left( x \right) +{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }\frac { n! }{ \left( n-2 \right) ! } \int _{ 0 }^{ c }{ { x }^{ n-2 }{ e }^{ -2x }dx } \right) } \end{aligned}

As you could see in the last step, the integral slightly changes, as well as the factor in front of it. But hopefully now, you see the pattern...

Now lets do this process $a$ times (where $n\ge a \ge 0$ and $a \in \mathbb{N}_{0}$). No I'm not going to describe it in detail (because it is the same as before; I also did it two times and otherwise, this note would get too large), thats something to pause and ponder. As before, we will store everything besides the integral (and its factors) in $Z \left( x \right)$: $\int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -2x }dx } =\lim _{ c\rightarrow \infty }{ \left( Z\left( x \right) +{ \left( \frac { 1 }{ 2 } \right) }^{ a }\frac { n! }{ \left( n-a \right) ! } \int _{ 0 }^{ c }{ { x }^{ n-a }{ e }^{ -2x }dx } \right) }$

And by repeating this process $n$ times, we can reduce the "integral of a product of functions" to an "integral of one function": $\int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -2x }dx } =\lim _{ c\rightarrow \infty }{ \left( Z\left( x \right) +{ \left( \frac { 1 }{ 2 } \right) }^{ n }n!\int _{ 0 }^{ c }{ { e }^{ -2x }dx } \right) }$ Now we can split the limit into two: $\int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -2x }dx } =\lim _{ c\rightarrow \infty }{ Z\left( x \right) } +\lim _{ c\rightarrow \infty }{ \left( { \left( \frac { 1 }{ 2 } \right) }^{ n }n!\int _{ 0 }^{ c }{ { e }^{ -2x }dx } \right) }$

But there is something important to mention! We can only do this algebraic "transformation" if both (created) limits exist. Its easy to check with the fundamental therorem of analysis that the right limit exists. \begin{aligned}\lim _{ c\rightarrow \infty }{ \left( { \left( \frac { 1 }{ 2 } \right) }^{ n }n!\int _{ 0 }^{ c }{ { e }^{ -2x }dx } \right) } &={ \left( \frac { 1 }{ 2 } \right) }^{ n }n!\lim _{ c\rightarrow \infty }{ \left( \int _{ 0 }^{ c }{ { e }^{ -2x }dx } \right) }\\ &={ \left( \frac { 1 }{ 2 } \right) }^{ n }n!\lim _{ c\rightarrow \infty }{ \left( { \left[ { \left( -\frac { 1 }{ 2 } \right) e }^{ -2x } \right] }_{ 0 }^{ c } \right) } \\&={ \left( \frac { 1 }{ 2 } \right) }^{ n }n!\lim _{ c\rightarrow \infty }{ \left( { \left( -\frac { 1 }{ 2 } \right) \frac { 1 }{ { e }^{ 2c } } }-{ \left( -\frac { 1 }{ 2 } \right) e }^{ -2\cdot 0 } \right) } \\&={ \left( \frac { 1 }{ 2 } \right) }^{ n }n!\frac { 1 }{ 2 } ={ \left( \frac { 1 }{ 2 } \right) }^{ n+1 }n!\end{aligned}

This is true, because $e^{-2c}$ is a null sequence.

But does the left limit exist as well? The answer is yes. At the end $Z$ is: $Z\left( x \right) ={ \left[ { x }^{ n }{ e }^{ -2x } \right] }_{ 0 }^{ c }+\frac { n }{ 2 } { \left[ { x }^{ n-1 }{ e }^{ -2x } \right] }_{ 0 }^{ c }+\frac { n\left( n-1 \right) }{ 4 } { \left[ { x }^{ n-2 }{ e }^{ -2x } \right] }_{ 0 }^{ c }+\dots +{ \left( \frac { 1 }{ 2 } \right) }^{ n }n!{ \left[ { x }{ e }^{ -2x } \right] }_{ 0 }^{ c }$ As you can see in every of these functions (in brackets) there is a power of $x$ and a $e^{-2x}$ term. If you plug the lower bound $0$ into these functions, the power of $x$ will cause every term to be zero. If you plug in the upper bound $c$, then the $e^{-2x}$ is a zero sequence and will cause every term to be zero. Thus we can say, that the whole limit (of $Z$) is equal to $0$.

Now we know, that both limits of $\lim _{ c\rightarrow \infty }{ Z\left( x \right) } +\lim _{ c\rightarrow \infty }{ \left( { \left( \frac { 1 }{ 2 } \right) }^{ n }n!\int _{ 0 }^{ c }{ { e }^{ -2x }dx } \right) }$ exist and therefore the split out is valid. And here are the last steps of this proof: \begin{aligned}\int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -2x }dx } &=\lim _{ c\rightarrow \infty }{ Z\left( x \right) } +\lim _{ c\rightarrow \infty }{ \left( { \left( \frac { 1 }{ 2 } \right) }^{ n }n!\int _{ 0 }^{ c }{ { e }^{ -2x }dx } \right) } \\&=0+{ \left( \frac { 1 }{ 2 } \right) }^{ n+1 }n!\\&={ \left( \frac { 1 }{ 2 } \right) }^{ n+1 }n!\end{aligned}

You can validate this integral very quick by induction. I could did it with induction, but for me this is in this context (in a node on brilliant.org) an odd method. Because induction does only proof it, and this way explains where the factorial term come from. Because I think, it is more important to explain a proof rather than just validate it. Note by CodeCrafter 1
1 year, 2 months ago

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## Comments

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This follows from the properties of the Gamma function: $\Gamma(z) = \int_0^\infty x^{z - 1} e^{-x} \ dx.$ For a positive integer $n$, $\Gamma(n) = (n - 1)!$. This means $\int_0^\infty x^n e^{-x} \ dx = n!.$ Then your integral follows by a simple substitution.

- 1 year, 2 months ago

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Interesting. I heared of the Gamma Function before. But I never looked deeper into this function. Now I will definitely do that.

And I find this multiple partial integral very funny ;)

- 1 year, 2 months ago

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