Evaluating the integral $\int_0^\infty \frac{\psi(x)-x}{x^2}dx.$

Consider $f(s):=\int_1^{\infty}\frac{\psi(x)-x}{x^s}dx$, which converges for $Re(s)\geq2$. For $Re(s)>2$, we can separate the numerator and integrate by parts (using the Riemann-Stieltjes integral, for convenience) to get $f(s)=\frac{1}{s-1}\int_1^{\infty}\frac{1}{x^{s-1}}d\psi(x)-\frac{1}{s-2}$. Now, $\frac{\zeta'}{\zeta}(s)=-\sum\frac{\Lambda(n)}{n^s}=-\int_1^{\infty}\frac{1}{x^s}d\psi(x)$, so we can write $f(s)=\frac{-1}{s-1}\frac{\zeta'}{\zeta}(s-1)-\frac{1}{s-2}$. $\frac{\zeta'}{\zeta}(s-1)$ has a Laurent expansion at $s=2$ of the form $\frac{\zeta'}{\zeta}(s-1)=\frac{-1}{s-2}+\gamma+O(s-2)$, so that $f(s)=\frac{-1-\gamma}{s-1}+O(s-2)$. This holds for $Re(s)>2$, but if we let $s$ decrease to 2 and use the dominated convergence theorem, we get that the value of the integral is $-\gamma-1$.

Hmm...this isn't quite the same as either value you gave. Maybe I made a mistake somewhere.

I want to prove it with an elementary approach:

a theorem of Landau say that the PNT is equivalent to $\sum_{n\leq x} \frac{\Lambda(n)}{n}=log(x)-\gamma+o(1)$.

Now using partial summation we have:

$\displaystyle \sum_{n\leq x} \frac{\Lambda(n)}{n}=\displaystyle \int_{1}^{x}\frac {d\psi(t)}{t}=\int_{1}^{x}\frac {d(\psi(t)-t)}{t}+\int_{1}^{x}\frac {dt}{t}=log(x)+\int_{1}^{+\infty}\frac {\psi(t)-t}{t^{2}}-\int_{x}^{+\infty}\frac {\psi(t)-t}{t^{2}}+\frac{\psi(x)-x}{x}-\frac{\psi(1)-1}{1}=log(x)+\int_{1}^{+\infty}\frac {\psi(t)-t}{t^{2}}+o(1)+1$

so that $\displaystyle \int_{1}^{+\infty}\frac {\psi(t)-t}{t^{2}}=-1-\gamma$.