A formula for $\int_{1}^{2}x^{-x}dx$?

Question

I can resist to ask here : It's a follow up of A new trick for the Sophomore's dream? :

Some definition :

\begin{eqnarray*} j(x)&=&\frac{c}{k}\sum_{n=1}^{k} ex^{\left(\frac{1}{k^{d}}+\frac{1}{k^{2}}\right)}e^{-x^{\left(1+\frac{1}{k^{d}}+\frac{1}{k^{2}}\right)}}\\ &&{}+\frac{\left(1-c\right)}{3k}\sum_{n=1}^{k} ex^{\left(\frac{1}{k^{2r}}+\frac{1}{k^{4}}\right)}e^{-x^{\left(1+\frac{1}{k^{2r}}+\frac{1}{k^{4}}\right)}}\\ &&{}+\frac{\left(1-c\right)}{3k}\sum_{n=1}^{k} ex^{\left(\frac{1}{k^{4}}+\frac{1}{k^{6}}\right)}e^{-x^{\left(1+\frac{1}{k^{4}}+\frac{1}{k^{6}}\right)}}\\ &&{}+\frac{\left(1-c\right)}{3k}\sum_{n=1}^{k} ex^{\left(\frac{1}{k^{6t}}+\frac{1}{k^{8}}\right)}e^{-x^{\left(1+\frac{1}{k^{6t}}+\frac{1}{k^{8}}\right)}} \end{eqnarray*}

Where

$$t=0.56,r=1.275,d=1.15,c=1.082,k=3$$

Then define :

\begin{eqnarray*} F(x)&=&\frac{\left|j(x)-x^{-x}\right|}{x}\\ a_{1}(x)&=&\frac{\left|F(x)-F(2)x(x-2)-xF(2)\right|}{x}\\ a_{2}(x)&=&a_{1}(1.5)(x-2)+a_{1}(1.5)\\ a_{3}(x)&=&\frac{\left|a_{1}(x)-xa_{2}(x)\right|}{x}\\ a_{4}(x)&=&a_{3}(1.5)(x-2)+a_{3}(1.5)\\ a_{5}(x)&=&\frac{\left|a_{3}(x)-x a_{4}(x)\right|}{x} \end{eqnarray*}

And so on…

I'm pretty sure that $a_n$ converges to zero on $(1,2)$.

The next thing is we can integrate easily the $a_n$.

So how to show that:

$$a_n\to 0,\ x\in[1,2]$$

What looks like the equality I mean:

$$\int_{1}^{2}x^{-x}dx=?$$

Answer 1

Is known, that $$(1+a)^{-b}=\sum\limits_{j=0}^\infty \dbinom{-b}{j}a^j,$$ $$(1+a)^{-b}=1-ba\left(1-(b+1)\dfrac a2\,\left(1-(b+2)\dfrac a3\,\left(1-\dots (b+k)\dfrac a{k+1}\dots\right)\right)\right).$$ At the same time, $$\int\limits_1^2 x^{-x}\,\text dx = \int\limits_1^{\large\frac32} x^{-x}\,\text dx +\int\limits_{\large\frac32}^2 x^{-x}\,\text dx.$$

If $\color{brown}{\mathbf {x\in\left[1,\frac32\right]}},$ then

$\qquad x=\dfrac54\left(1+\dfrac y5\right)=\dfrac54+\dfrac{y}4,\quad$ where $y\in[-1,1].$

Therefore, $$x^{-x} = \left(\dfrac54\right)^{\Large-\frac{y+5}4}\left(1+\dfrac y5\right)^{\large-\frac54}\left(1+\dfrac y5\right)^{\Large-\frac{y}4},$$ wherein $$\left(1+\dfrac y5\right)^{\Large -\frac {y}4}\approx P_{16}(y)= 1-\dfrac{y^2}{20}\bigg(1-\dfrac{y^2+4y}{40}\,\bigg(1-\dfrac{y^2+8y}{60}\, \bigg(1-\dfrac{y^2+12y}{80}\,$$ $$\times\left(1-\dfrac{y^2+16y}{100}\,\left(1-\dfrac{y^2+20y}{120}\,\left(1-\dfrac{y^2+24y}{140}\,\left(1-\dfrac{y^2+28y}{160}\,\right)\right)\right)\right)\bigg)\bigg)\bigg)$$ Let $y=5z-5,$ then $x=\dfrac54 z,\quad$ and $$I_1=\int\limits_1^{\large\frac32} x^{-x}\,\text dx\approx \int\limits_{\large\frac45}^{\large\frac65}\left(\dfrac54\right)^{\large-\frac54\,z-1}z^{\large-\frac54} P_{16}(5z-5)\,\text dz.$$ Using of the integral $$\int\limits_{\large\frac45}^{\large\frac65}\left(\dfrac54\right)^{\large-\frac54\,z-1}z^{k-\large\frac54}\,\text dz =\left(\dfrac54\right)^{\large\frac54-k} E_{\large\frac54-k} \left(\ln\,\frac54\right) - 2^{\large k-\frac94} 3^{\large k-\frac14} 5^{\large\frac54-k}E_{\large\frac54-k}\left(\ln\dfrac{5\sqrt5}8\right),$$ where $E_n(x)$ is the exponential integral, leads to the result $$I_1\approx 0.3809511001490732.$$ Numeric integration leads to the result $$I_1\approx 0.3809510996568984.$$

If $\color{brown}{\mathbf {x\in\left[\frac32,2\right]}},$ then

$\qquad x=\dfrac74\left(1+\dfrac y7\right)=\dfrac74+\dfrac{y}4,\quad$ where $y\in[-1,1].$

Therefore, $$x^{-x} = \left(\dfrac74\right)^{\Large-\frac{y+7}4}\left(1+\dfrac y7\right)^{\large-\frac74}\left(1+\dfrac y7\right)^{\Large-\frac{y}4},$$ wherein $$\left(1+\dfrac y7\right)^{\Large -\frac {y}4}\approx Q_{16}(y)= 1-\dfrac{y^2}{28}\bigg(1-\dfrac{y^2+4y}{56}\,\bigg(1-\dfrac{y^2+8y}{84}\, \bigg(1-\dfrac{y^2+12y}{112}\,$$ $$\times\left(1-\dfrac{y^2+16y}{140}\,\left(1-\dfrac{y^2+20y}{168}\,\left(1-\dfrac{y^2+24y}{196}\,\left(1-\dfrac{y^2+28y}{224}\, \right)\right)\right)\right)\bigg)\bigg)\bigg)$$ Let $y=7z-7,$ then $x=\dfrac74 z,\quad$ and $$I_2=\int\limits_{\large\frac32}^2 x^{-x}\,\text dx\approx \int\limits_{\large\frac67}^{\large\frac87}\left(\dfrac74\right)^{\large-\frac74\,z-1}z^{\large-\frac74} Q_{16}(7z-7)\,\text dz.$$ Using of the integral $$\int\limits_{\large\frac67}^{\large\frac87}\left(\dfrac74\right)^{\large-\frac74\,z-1}z^{k-\large\frac74}\,\text dz =2^{\large k-\frac{11}4} 7^{\large\frac74-k} \left(3^{\large k-\frac34} E_{\large\frac74 - k}\left(\ln\,\frac{7\sqrt7}8\right) - 2^{\large 2k-\frac32} E_{\large\frac74-k}\left(\ln\,\frac{49}{16}\right)\right)$$ leads to the result $$I_2\approx 0.1913981347868081.$$ Numeric integration leads to the result $$I_2\approx 0.1913981347751783.$$

This allows to get $$I\approx 0.5723492349358814,$$ when the numeric calculations give $$I\approx 0.5723492344320767.$$

Easily to see that the proposed method provides the arbitrary accuracy.