# Integral involving $u^{-u}$

I'm trying to solve the following definite integral

$$\int_0^{\infty}u^{-u+a}\,du$$ with $$a>0$$.

Using integration by parts I've arrived

$$\int_0^{\infty}u^{-u+a}\,du=\frac{1}{1+a}\int_0^{\infty}u^{-u+a+1}(1+\ln(u))\,du$$

which is a worst integral.

Maybe a change of variable, but I don't know how to follow.

• I'm afraid there is no elementary integral for $u^{-u}$ – Bernard Massé Jan 31 at 1:26
• Do you wanna say that it's not possible express in terms of elementary functions? And...in terms of special functions? – popi Jan 31 at 1:28
• Notice that the minimum value of the integral correspond to $a \approx \log(2)$. Why ? This is a question. – Claude Leibovici Jan 31 at 6:07

Let $$I_a=\int_0^{\infty}u^{-u+a}\,du$$
For small values of $$a$$, you would get $$\left( \begin{array}{cc} a & I_a \\ 0.00 & 1.99546 \\ 0.25 & 1.81433 \\ 0.50 & 1.73133 \\ 0.75 & 1.71513 \\ 1.00 & 1.75183 \\ 1.25 & 1.83622 \\ 1.50 & 1.96842 \\ 1.75 & 2.15247 \\ 2.00 & 2.39609 \end{array} \right)$$
For large values of $$a$$, the integral varies extremely fast $$\left( \begin{array}{cc} a & I_a \\ 0 & 1.99546 \\ 1 & 1.75183 \\ 2 & 2.39609 \\ 3 & 4.27169 \\ 4 & 9.22902 \\ 5 & 23.2062 \\ 6 & 66.1712 \\ 7 & 210.120 \\ 8 & 733.083 \\ 9 & 2781.12 \\ 10 & 11378.2 \end{array} \right)$$