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

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.

• 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. – user37238 Aug 21 '15 at 7:22
• 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.) – Raymond Manzoni Aug 21 '15 at 7:33

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

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}

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, $$1+2+3=\int_{0}^{\infty}t^3e^{-t}dt$$$$\iff 1+2+3=3!$$$$\iff 6=6$$

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.

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.