1. ## Improper integrals:

$\int_{\frac{1}{2}}^{2}\frac{dz}{z(ln{z})^{\frac{1} {5}}}$

so:

$\lim_{t \to 1^{-}} \int_{\frac{1}{2}}^{t} \frac{dz}{z(ln{z})^{\frac{1}{5}}} + \lim_{t\to 1^{+}} \int_{t}^{2}\frac{dz}{z(ln{z})^{\frac{1}{5}}}$

$\lim_{t\to{1^{-}}} \frac{5}{4} (ln{z})^{\frac{4}{5}} \big|_{\frac{1}{2}}^{t} + \lim_{t\to{1^{+}}} \frac{5}{4} (ln{z})^{\frac{4}{5}}\big|_{t}^{2}$

when i evaluate this.... the answer doesnt appear to be the sam as the back of book:
Answer at back of book : 0

$\int_{\frac{1}{2}}^{2}\frac{dz}{z(ln{z})^{\frac{1} {5}}}$

so:

$\lim_{t \to 1^{-}} \int_{\frac{1}{2}}^{t} \frac{dz}{z(ln{z})^{\frac{1}{5}}} + \lim_{t\to 1^{+}} \int_{t}^{2}\frac{dz}{z(ln{z})^{\frac{1}{5}}}$

$\lim_{t\to{1^{-}}} \frac{5}{4} (ln{z})^{\frac{4}{5}} \big|_{\frac{1}{2}}^{t} + \lim_{t\to{1^{+}}} \frac{5}{4} (ln{z})^{\frac{4}{5}}\big|_{t}^{2}$

when i evaluate this.... the answer doesnt appear to be the sam as the back of book:
Answer at back of book : 0
$\lim_{t\to{1^{-}}} \frac{5}{4} (ln{z})^{\frac{4}{5}} \big|_{\frac{1}{2}}^{t} + \lim_{t\to{1^{+}}} \frac{5}{4} (ln{z})^{\frac{4}{5}}\big|_{t}^{2} =$ $\lim_{t \to 1} \left[ \frac {5}{4} (\ln t )^{4/5} - \frac {5}{4}( - \ln 2)^{4/5} \right] + \lim_{t \to 1} \left[ \frac {5}{4} (\ln 2)^{4/5} - \frac {5}{4}( \ln t)^{4/5} \right]$

$= - \frac {5}{4} (-\ln 2)^{4/5} + \frac {5}{4}(\ln 2)^{4/5}$

$= - \frac {5}{4} \left[ (\ln 2)^4 \right]^{1/5} + \frac {5}{4} \left[ (\ln 2)^4 \right]^{1/5}$

$= 0$

3. thanks:
i got it now so $\ln{\frac{1}{2}} = \ln{1} - \ln{2} = -\ln{2}$

i got it now so $\ln{\frac{1}{2}} = \ln{1} - \ln{2} = -\ln{2}$
Or $ln \left ( \frac{1}{2} \right ) = ln \left ( 2^{-1} \right ) = -ln(2)$