4
$\begingroup$

I'm reading Ivanov's: Easy as Pi. In the cover of the book, there is a formula:

$$1+2+3=\int_{0}^{\infty}t^3e^{-t} dt$$

It's not clear to me if the formula has any relevance or if it is a joke. I skimmed through the book but didn't find anything related to that in a direct way.

$\endgroup$
  • $\begingroup$ As pointed out in Ross Millikan's answer, the interesting question is how to make the $1+2+3$ appear as steps in the derivation. $\endgroup$ – user37238 Aug 21 '15 at 7:22
  • $\begingroup$ I find $\;1\cdot 2\cdot 3\cdots n=\int_{0}^{\infty}t^n\,e^{-t} dt\,$ somewhat nicer... (possible typo on the cover from the russian trans.) $\endgroup$ – Raymond Manzoni Aug 21 '15 at 7:33
5
$\begingroup$

The right side is the Gamma function. It is equivalent to $3!=6=1+2+3$

See Gamma Function: https://en.wikipedia.org/wiki/Gamma_function

$\endgroup$
6
$\begingroup$

You may just integrate by parts repeatedly.

Here are the details.

$$ \begin{align} \int_0^{\infty}t^3e^{-t}dt&=\left. t^3\left(-e^{-t}\right)\right|_0^{\infty} +3\int_0^{\infty}t^2e^{-t}dt\\\\ &=0+3\times\int_0^{\infty}t^2e^{-t}dt\\\\ &=3\times\left(\left. t^2\left(-e^{-t}\right)\right|_0^{\infty} +2\int_0^{\infty}te^{-t}dt\right)\\\\ &=3\times\left(0+2\int_0^{\infty}te^{-t}dt\right)\\\\ &=3\times2 \times\int_0^{\infty}te^{-t}dt\\\\ &=3\times2\times\left(\left. t\left(-e^{-t}\right)\right|_0^{\infty} +\int_0^{\infty}e^{-t}dt\right)\\\\ &=3\times2\times\left(0 -(-1)\right)\\\\ &=3\times2\times1\\\\ &=\color{blue}{6}\\\\ &=\color{red}{1+2+3}. \end{align} $$

$\endgroup$
4
$\begingroup$

Notice, $$RHS=\int_{0}^{\infty}t^3e^{-t}dt=L[t^3]_{s=1}=\left[\frac{\Gamma(3+1)}{s^{3+1}}\right]_{s=1}=\left[\frac{\Gamma(4)}{1^{4}}\right]=\Gamma(4)=3!=3\times2\times 1=6$$ Hence, $$1+2+3=\int_{0}^{\infty}t^3e^{-t}dt$$$$\iff 1+2+3=3!$$$$\iff 6=6$$

$\endgroup$
4
$\begingroup$

You can derive it by integrating by parts. If $u=t^3, dv=e^{-t}dt, \int_{0}^{\infty}t^3e^{-t} dt=-t^3e^{-t}|_0^\infty+\int_0^\infty3t^2e^{-t}dt\to\int_0^\infty6e^{-t}dt=6=1+2+3$. I don't know how to make the $1+2+3$ appear as steps in the derivation.

$\endgroup$
2
$\begingroup$

By definition $$I_n=\int t^n e^{-t}\,dt=-\Gamma (n+1,t)$$ where appears the incomplete gamma function.$$J_n=\int_0^\infty t^n e^{-t}\,dt=\Gamma (n+1)$$ provided that $n>-1$. If $n$ is an integer, then $$J_n=n!$$ The case of $n=3$ is then very particular.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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