0
$\begingroup$

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.

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

This looks even worse than the sophomore's dream and only numerical methods could be used.

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 $$\left( \begin{array}{cc} a & I_a \\ 0 & 1.99546 \\ 1 & 1.75183 \\ 2 & 2.39609 \\ 3 & 4.27169 \\ 4 & 9.22902 \\ 5 & 23.2062 \\ 6 & 66.1712 \\ 7 & 210.120 \\ 8 & 733.083 \\ 9 & 2781.12 \\ 10 & 11378.2 \end{array} \right)$$

$\endgroup$

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.