# Thread: How can i explain???

1. ## How can i explain???

I have shown that:

When $\rho = \beta + i \gamma$

Then

$\frac{x^{\rho}}{\rho} +\frac{x^{\overline \rho}}{\overline\rho} = \frac{2x^{\beta}}{|\rho|} cos(\gamma In(x) - \theta)$

However I don't how I can explain why:

$Li(x^{\rho}) \sim \frac{x^{\rho}}{\rho Inx}$

2. Originally Posted by signature
I have shown that:

When $\rho = \beta + i \gamma$

Then

$\frac{x^{\rho}}{\rho} +\frac{x^{\overline \rho}}{\overline\rho} = \frac{2x^{\beta}}{|\rho|} cos(\gamma In(x) - \theta)$

However I don't how I can explain why:

$Li(x^{\rho}) \sim \frac{x^{\rho}}{\rho Inx}$
First note that $\frac{x^\rho}{\rho\log(x)}=\frac{x^\rho}{\log(x^\r ho)}$.

Let $t=x^\rho$. Now integrate by parts: $\text{li}(t)=\int_2^t\frac{1}{\log(t)}dt=\frac{t}{ \log(t)}-\frac{2}{\log(2)}+\int_2^t\frac{1}{\log^2(t)}dt$.

What's left to show is $\lim_{t\to\infty} \frac{\int_2^t\frac{1}{\log^2(t)}dt}{\frac{t}{\log (t)}}=0$. I'll let you do it... apply L'hopital's rule to get it.

Thus $\lim_{t\to\infty} \frac{\text{li}(t)}{\frac{t}{\log(t)}}=1 \implies \text{li}(t)\sim\frac{t}{\log(t)}$