Following On the integral $I_s =\int_{1}^{\infty} (\pi(x)-Li(x))x^{-s-1} dx$, define

$$I_{s}=\int_{1}^{\infty} (\pi(x)-Li(x))x^{-s-1} \mathrm{d}x,$$ where $\pi$ denotes the prime counting function and $Li$ the logarithmic integral.

My question is, does $I_s$ converge at $s=\Theta$, where $\Theta$ is the minimal real number such that $\pi(x)-Li(x) \ll x^{\Theta+ \epsilon}$ for any $\epsilon>0$ ?

Following On the integral $I_s =\int_{1}^{\infty} (\pi(x)-Li(x))x^{-s-1} dx$, define

$$I_{s}=\int_{1}^{\infty} (\pi(x)-Li(x))x^{-s-1} \mathrm{d}x,$$ where $\pi$ denotes the prime counting function and $Li$ the logarithmic integral.

Does $I_s$ converge at $s=\Theta$, where $\Theta$ is the minimal real number such that $\pi(x)-Li(x) \ll x^{\Theta+ \epsilon}$ for any $\epsilon>0$ ?

Following On the integral $I_s =\int_{1}^{\infty} (\pi(x)-Li(x))x^{-s-1} dx$, define

$$I_{s}=\int_{1}^{\infty} (\pi(x)-Li(x))x^{-s-1} \mathrm{d}x,$$ where $\pi$ denotes the prime counting function and $Li$ the logarithmic integral.

My question is, does $I_s$ converge at $s=\Theta$, where $\Theta$ is the minimal number such that $\pi(x)-Li(x) \ll x^{\Theta+ \epsilon}$ for any $\epsilon>0$ ?

Following On the integral $I_s =\int_{1}^{\infty} (\pi(x)-Li(x))x^{-s-1} dx$, define

$$I_{s}=\int_{1}^{\infty} (\pi(x)-Li(x))x^{-s-1} \mathrm{d}x,$$ where $\pi$ denotes the prime counting function and $Li$ the logarithmic integral.

My question is, does $I_s$ converge at $s=\Theta$, where $\Theta$ is the minimal real number such that $\pi(x)-Li(x) \ll x^{\Theta+ \epsilon}$ for any $\epsilon>0$ ?