I was trying to find whether the $\int_{0}^\infty\frac{\ln(x\pi^x)\,dx}{x}$ converges or not. And after integrating I got
$\lim_{x\to\infty}(\frac{1}{2}\ln^2(x)+x\ln\pi)-\lim_{x\to0}(\frac{1}{2}\ln^2(x)+x\ln\pi)$. (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, $$\lim_{x\to\infty}\left[ \left(\frac{1}{2}\ln^2(x)+x\ln\pi\right) - \left(\frac{1}{2}\ln^2(1/x)+(1/x)\ln\pi\right)\right]$$ (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:
- Does A=B (doesn't seem so)
- Does the integral converges or diverges? (And how to figure that out)
We have that
$$\int_{0}^\infty\frac{\ln(x\pi^x)}{x}dx =\int_{0}^\infty\frac{\ln x +x\ln \pi }{x}dx =\int_{0}^\infty\frac{\ln x}{x}dx+\int_{0}^\infty\ln \pi \,dx$$
and we can conclude here since both integrals diverge.
Note that for the limit we have
$$\left(\frac{1}{2}\ln^2(x)+x\ln\pi\right) - \left(\frac{1}{2}\ln^2(1/x)+(1/x)\ln\pi\right)$$
$$\frac{1}{2}\ln^2(x)+x\ln\pi - \frac{1}{2}\ln^2(1/x)-(1/x)\ln\pi$$
$$\frac{1}{2}\ln^2(x)+x\ln\pi - \frac{1}{2}\ln^2(x)-(1/x)\ln\pi$$
$$x\ln\pi -(1/x)\ln\pi \to \infty$$
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
$$\int_{0}^\infty\frac{\ln(x\pi^x)}{x}dx=\int_{0}^a \frac{\ln(x\pi^x)}{x}dx+\int_{a}^\infty\frac{\ln(x\pi^x)}{x}dx$$
and the two limits at $0$ and $\infty$ for the two integrals are independent and must be evaluated separetely.