Show by hand $\int_{0}^{\sqrt{5}}x^{x}dx>4$

Question

It's a very Challenging question perhaps a kind of olympiad question :

Prove that :

$$\int_{0}^{\sqrt{5}}x^{x}dx>4$$

It's pretty sharp since the left hand side is almost $4.0005$ .

I recall the Sophomore dream :

$$\int_{0}^{1}x^{x}dx=-\sum_{n=1}^{\infty}\left(-n\right)^{-n}$$

For the other part we have :

$$\int_{1}^{\sqrt{5}}x^{x}dx\geq \int_{1}^{\sqrt{5}}\left(1-\sqrt{x}+x\right)^{1+x}dx$$

Because I show in another question the inequality $x>0$:

$$\left(1-\sqrt{x}+x\right)^{1+x}\leq x^x$$

But we are far from the goal .

We have numerically :

$$\int_{1}^{1.5}x^{x}dx>\int_{1}^{1.5}\left(e^{\left(-1+\frac{\left(1+x+x^{2}\right)}{3}\right)}+\frac{1}{6}\left(x-1\right)^{2}+\frac{1}{26}\left(x-1\right)^{4}\right)dx$$

How to show it by hand or make a proof where the final steps is the calculus of some constant ?

Answer 1

$\color{green}{\textbf{Splitting.}}$

Let $\;L=\dfrac{582651}{582601},\quad q=\dfrac{44}{43},\;$ then $\;L<q^{-35}\sqrt 5,$ $$I>\int\limits_0^{\sqrt5} x^x\,\text{dx} >\int\limits_0^1 x^x\,\text dx+\int\limits_1^L x^x\,\text dx+\sum_{k=0}^{34}\int\limits_{q^{k-35}\sqrt5}^{q^{k-34}\sqrt5} x^x\,\text dx.\tag1$$

$\color{green}{\textbf{First integral.}}$

In accordance with OP, $$I_0=\int_0^1 x^x\,\text dx = \sum_{k=1}^\infty (-1)^{k+1} k^{-k},\tag{$\Diamond$}$$ so $$I_0 > \sum_{k=1}^8 (-1)^{k+1} k^{-k} >\dfrac1{8^8}\sum_{k=1}^8 \genfrac\lfloor\rfloor{}{}{(-1)^{k+1}8^8}{k^k}=\dfrac1{16777216}(16777216-4194304+621378-262144+5368-359+20-1),$$ $$I_0=\dfrac{13143781}{16777216}>0.783430.\tag2$$

$\color{green}{\textbf{Second interal.}}$

$$I_1=\int\limits_1^L x^x\,\text dx = \int\limits_0^{L-1} (1+y)^{1+y}\,\text dy \ge \int_0^{L-1} (1+y(1+y)) \,\text dy = \dfrac{(L-1)(6+3(L-1)+2(L-1)^2)}6,$$ $$I_1=\dfrac{50(6\cdot 582601^2+3\cdot 50\cdot 582601+2\cdot 50^2)}{6\cdot582601^3} > 0.0000858.\tag3$$

$\color{green}{\textbf{Integrals under the sum.}}$

Let $x\in\big(a,a+\delta a\big),\;\delta=q-1=\dfrac1{43}\ll1,\;$ then $$x=a(1+\delta z)=a +\delta a z,\qquad \big(z\in(0,1),\;a\in(1,\infty)\big),$$ $$x^{x}=a^{\large a+\delta a z}\;\left(1+\delta z \right)^{\large a(1+\delta z)},$$

Taking in account the inequality $$(1+t)^{\large a(1+t)}- e^{at} \ge \dfrac a2 t^2\left(1+\dfrac{3a-1}3 t\right)\tag{4}$$ and identities

$$\int\limits_0^d t^k a^{a t}\,\text dt = \dfrac{\Gamma(1+k)-\Gamma(1+k,-a d\ln(a))}{(-a \ln(a))^{1+k}},$$ $$\dfrac{\Gamma(k+1)-\Gamma(k+1,-y)}{(-y)^{k+1}} =\dfrac1{k+1}+\dfrac y{k+2}+\dfrac{y^2}{2!(k+3)}+\dots,$$ one can get $$\int\limits_0^d t^k a^{a t}\,\text dt = \dfrac{\Gamma(1+k)-\Gamma(1+k,-ad\ln(a))}{(-a d\ln(a))^{k+1}} \dfrac{(ad\ln a)^{k+1}}{(-a \ln(a))^{k+1}},$$ $$\int\limits_0^d t^k a^{a t}\,\text dt =d^{k+1}\left(\dfrac1{k+1}+\dfrac {a d\ln(a)}{k+2}+\dfrac{(ad\ln(a))^2}{2!(k+3)}+\dots\right),\tag5$$ $$\begin{align} &J(a, \delta)=\int\limits_a^{a+\delta a} x^x\,\text dx = \delta a\cdot a^a \int\limits_0^1 a^{\delta az}\left(1+\delta z\right)^{\large a(1+\delta z)}\,\text dz\\[4pt] &= a^{1+a}\int\limits_0^\delta a^{at}\left(1+t\right)^{\large a(1+t)}\,\text dt \ge a^{1+a} \int\limits_0^\delta a^{\large at} \left(e^{at}+\frac a2 t^2 + \frac {a(3a-1)}6 t^3\right)\,\text dt\\[4pt] &=a^{1+a} \int\limits_0^\delta (ea)^{\large a t}\,\text dt + \dfrac12a^{2+a} \int\limits_0^\delta t^2 a^{\large a t}\,\text dt +\dfrac {3a-1}6 a^{2+a} \int\limits_0^\delta t^3 a^{\large a t} \,\text dt,\\[4pt] &J(a,\delta)\ge a^a\dfrac{(ea)^{\large \delta a}-1}{\ln(ea)} +\dfrac12a^{a+2}\delta^3\left(\dfrac13+\dfrac {a\delta\ln(a)}4 +\dfrac{(a\delta\ln(a))^2}{10}+\dots\right)\\[4pt] &+\dfrac{3a-1}6 a^{a+2}\delta^4\left(\dfrac14+\dfrac {a\delta\ln(a)}5+\dfrac{(a\delta\ln(a))^2}{12}+\dots\right), \end{align}$$ $$J\left(a,\dfrac1{43}\right) > a^a\left(\dfrac{(ea)^{\large \frac a{43}}-1}{\ln(ea)}+\dfrac{a^2}{6\cdot 43^3}\left(1+\dfrac{3a-1}{196}\right)\right).\tag6$$ Therefore, $$I_2>\sum_{k=1}^{35} J\left(\left(\dfrac{43}{44}\right)^k\,\sqrt5, \dfrac1{43}\right) > 3.216991.\tag7$$

$\color{green}{\textbf{Final summation.}}$

Taking in account $(1),$ $(2),$ $(3),$ $(7),$ easily to get

$$\color{brown}{\mathbf{\int\limits_0^{\sqrt5} x^x\,\text dx> I_0+I_1+I_2 > 0.783430+ 0.0000858+3.216991 = 4.000506>4.}}$$

Note, that calculations contain only elementary functions and can be executed on the calculator.

Answer 2

We may use the generalized binomial theorem to get a result with arbitrary precision. We start with $$\int_{0}^{1}x^{x}dx=\sum_{n\geq1}\frac{\left(-1\right)^{n+1}}{n^{n}}$$ and, obviously, if we fix an even number $N_{1}\in\mathbb{N}$ we get $$\int_{0}^{1}x^{x}dx>\sum_{n\le N_{1}}\frac{\left(-1\right)^{n+1}}{n^{n}}.$$ Then we fix an odd number $N_{2}\in\mathbb{N}_{>2}$ and we consider $$\int_{1}^{2}x^{x}dx=\int_{1}^{2}\left(x-1+1\right)^{x}dx>\sum_{k\leq N_{2}}\frac{1}{k!}\int_{1}^{2}x\left(x-1\right)\cdots\left(x-k+1\right)\left(x-1\right)^{k}dx.$$ Note that all integrals are very simple to compute. Lastly, we take an even $N_{3}\in\mathbb{N}_{>2}$ and we consider $$\int_{2}^{\sqrt{5}}x^{x}dx>\sum_{k\leq N_{3}}\frac{2^{-k}}{k!}\int_{2}^{\sqrt{5}}x\left(x-1\right)\cdots\left(x-k+1\right)2^{x}\left(x-2\right)^{k}dx$$ and observe that the integrals can be easily computed using integration by parts, so we can obtain an error of approximation with arbitrary precision. For example, if we take $N_{1}=4,\,N_{2}=5,\,N_{3}=4$ we get $$\int_{0}^{\sqrt{5}}x^{x}dx>0.783130787+2.050245+1.1666425=4.000018287.$$

Answer 3

Let us get the accurate solution. $$\begin{align} &I=\int\limits_0^{\sqrt5} x^{x}\,\text dx = \sqrt5\int\limits_0^1 (x\sqrt5)^{x\sqrt5}\,\text dx = \sqrt5\int\limits_0^1 e^{x\sqrt5(\ln\sqrt5+\ln x)}\,\text dx\\[4pt] &=\sqrt5\int\limits_0^1\,\sum\limits\limits_{k=0}^\infty \dfrac{(x\sqrt5(\ln\sqrt5+\ln x))^k}{k!}\,\text dx = \sum\limits_{k=0}^\infty \dfrac{\left(\sqrt5\ln\sqrt5\right)^{k+1}}{k!} \int\limits_0^1\, x^k\left(1+\dfrac{\ln x}{\ln \sqrt5}\right)^k\,\text dx\\[4pt] &= \sum\limits_{k=0}^\infty \dfrac{\left(\sqrt5\right)^{k+1}}{k!} \sum\limits_{j=0}^k\dbinom kj\left(\ln\sqrt5\right)^{k+1-j} \int\limits_0^1\, x^k\left(\ln x\right)^j\,\text dx\\[4pt] &=\sum\limits_{k=0}^\infty \dfrac{\left(\sqrt5\right)^{k+1}}{k!} \sum\limits_{j=0}^k\dbinom kj \left(\ln\sqrt5\right)^{k+1-j} \left(k+1\right)^{-(j+1)} \int\limits_0^1\, \left(\ln\left(x^{k+1}\right) \right)^j\,\text d\left(x^{k+1}\right)\\[4pt] &=\sum_{k=0}^\infty \dfrac{\left(\sqrt5\right)^{k+1}}{k!} \sum\limits_{j=0}^k\dbinom kj \left(\ln\sqrt5\right)^{k+1-j} \left(k+1\right)^{-(j+1)} \int\limits_0^1\, \ln^j\left(y\right)\,\text dy\\[4pt] &=\sum\limits_{k=0}^\infty \dfrac{\left(\sqrt5\right)^{k+1}}{k!} \sum\limits_{j=0}^k \dfrac{k!}{j!(k-j)!} \left( \ln\sqrt5\right)^{k+1-j}(k+1)^{-(j+1)} (-1)^j j!\\[4pt] &=\sum\limits_{k=1}^\infty \left(\sqrt5\right)^{k+1} \sum\limits_{j=0}^k\left(\ln\sqrt5\right)^{k+1-j}\dfrac{(-1)^j}{(k-j)!(k+1)^{j+1}}\\[4pt] &=\ln\sqrt5\sum\limits_{k=0}^\infty (-1)^k \dfrac{\left(\sqrt5\right)^{k+1}}{(k+1)^{k+1}} \sum_{j=0}^k \dfrac{(-1)^{k-j}}{(k-j)!}\left(\ln\sqrt5\right)^{k-j}(k+1)^{k-j}\\[4pt] &=\ln\sqrt5\sum\limits_{k=1}^\infty (-1)^k \dfrac{\left(\sqrt5\right)^{k+1}}{(k+1)^{k+1}} \sum\limits_{j=0}^k \dfrac{(-1)^j}{j!}\left(\ln\sqrt5\right)^j(k+1)^j\\[4pt] &=\ln\sqrt5\sum\limits_{k=0}^\infty (-1)^k \dfrac{\left(\sqrt5\right)^{k+1}}{(k+1)^{k+1}} \dfrac{\Gamma(k+1, -(k+1)\ln\sqrt5)}{\left(\sqrt5\right)^{k+1} k!}, \end{align}$$ $$I=\ln\sqrt5\sum\limits_{k=1}^\infty \dfrac{(-1)^{k+1}\,\Gamma(k, -k\ln\sqrt5)} {k^k \Gamma(k)}.\tag1$$ Denote $$I_n=\ln\sqrt5\sum_{k=1}^n\limits\dfrac{(-1)^{k+1}\,\Gamma(k, -k\ln\sqrt5)} {{k^k} \Gamma(k)}.\tag2$$ Then all terms under sum are positive, wherein $$\begin{align} &I_8\ge 4.00014255542856536601434755919436288491029289677>4,\\[4pt] &I_{16}\ge 4.00051936486546223791324050775330864840365750833,\\[4pt] &I_{32}-I_{16}\ge 3.81485620187357894881186558671472983561471432051× 10^{-11}. \end{align}$$

Therefore,

• $I>I_8>4,$ i.e. the given inequality is proved;
• integral $\,I\,$ can be calculated with the arbitrary accuracy.