Vtyjkyuk

I was trying to find whether the $int_0^inftyfracln(xpi^x),dxx$ converges or not. And after integrating I got

$lim_xtoinfty(frac12ln^2(x)+xlnpi)-lim_xto0(frac12ln^2(x)+xlnpi)$. (Which I will define as A)

and I was lost there. Because I got $infty$-$infty$. Does that mean the integral diverges to something that is indeterminable.(not $infty$ and -$infty$)

Also I asked a question concerning the expression above here
L'hospital for inf-inf

And found that according to a fellow name gimusi, $$lim_xtoinftyleft[ left(frac12ln^2(x)+xlnpiright) - left(frac12ln^2(1/x)+(1/x)lnpiright)right]$$ (which I will define as B)

is, however calculable. And I calculated it to be approaching infinity. Does that mean the integral diverges (to infinity)? I am lost now.

Questions Summary:

1. Does A=B (doesn't seem so)


2. Does the integral converges or diverges? (And how to figure that out)

We have that

$$int_0^inftyfracln(xpi^x)xdx
=int_0^inftyfracln x +xln pi xdx
=int_0^inftyfracln xxdx+int_0^inftyln pi ,dx$$

and we can conclude here since both integrals diverge.

Note that for the limit we have

$$left(frac12ln^2(x)+xlnpiright) - left(frac12ln^2(1/x)+(1/x)lnpiright)$$

$$frac12ln^2(x)+xlnpi - frac12ln^2(1/x)-(1/x)lnpi$$

$$frac12ln^2(x)+xlnpi - frac12ln^2(x)-(1/x)lnpi$$

$$xlnpi -(1/x)lnpi to infty$$

but it can't be used to evaluate the integral indeed for the improper integral we need to consider for $a>0$ the two integrals

$$int_0^inftyfracln(xpi^x)xdx=int_0^a fracln(xpi^x)xdx+int_a^inftyfracln(xpi^x)xdx$$

and the two limits at $0$ and $infty$ for the two integrals are independent and must be evaluated separetely.