I can't solve following exercise in a note about prime numbers. I need this for study about large gaps of consecutive prime numbers.


Prove that f $0<1-\delta<1$ then


$$\sum\_{p\le y}\frac{1}{p^{1-\delta}}\le\frac{y^\delta}{\log(y^\delta)}+\log(1/\delta)+O\left(\frac{y^\delta}{\delta(\log y)^2}+1\right)$$.


In the note, a hint was given, that is "Split the sum at $p\le e^{2/\delta}$".
How this estimate can be proven?



An estimate of the sum related to primes
I only get  

$$\sum\_{p\le y}\frac{1}{p^{1-\delta}}\le \frac{y^{\delta}}{\log(y^\delta)}
+e^2\log(1/\delta)+O\Bigl(\frac{y^\delta}{\delta^2(\log y)^2}+1\Bigr).$$


The first sum is
$$\sum\_{p\le e^{2/\delta}}\frac{1}{p^{1-\delta}}=
\sum\_{p\le e^{2/\delta}}\frac{p^\delta}{p}\le e^2
\sum\_{p\le e^{2/\delta}}\frac{1}{p}$$
and applying the known bound on the sum of reciprocal of primes we get
$$e^2\log\log e^{2/\delta}+ C+O{(\log x)^{-1}}= e^2\log(1/\delta)+O(1).$$


For the second sum for $2^{1/\delta}<p\le y$ we have


$$=
\frac{\pi(y)}{y^{1-\delta}}-
\frac{\pi(2^{1/\delta})}{2^{(1-\delta)/\delta}}+(1-\delta)\int\_{2^{1/\delta}}^y\pi(t)
t^{-2+\delta}\,dt\le$$
$$\le \frac{y^\delta}{y}\Bigl(\frac{y}{\log y}+O(y/(\log y)^2)\Bigr)+
(1-\delta)\int\_{2^{1/\delta}}^y\frac{t}{\log t}
t^{-2+\delta}\,dt+$$
$$O\Bigl(\int\_{2^{1/\delta}}^y\frac{t}{(\log t)^2}
t^{-2+\delta}\,dt\Bigr)=$$
$$\le\frac{y^{\delta}}{\log y}+O\Bigl(\frac{y^{\delta}}{(\log y)^2}\Bigr)+
\frac{1-\delta}{\delta}\frac{y^\delta}{\log y}+
O\Bigl(\frac{y^\delta}{\delta^2(\log y)^2}\Bigr).$$


