How to evaluate the integral $\int_1^n\frac{1}{(\ln x)^{\ln x}}dx$ I'm stuck with this integral to test the convergence of a series. $$\int_1^n\frac{1}{(\ln x)^{\ln x}}dx.$$ Could you give me a couple of hints to compute this integral please? Is it a simple integral or I need to know something special to solve it? 

\begin{array}{l}
\int {\left( {{{\left( {\ln x} \right)}^{ - \ln x}}} \right)} \;dx\\
 = \int {\left( {{{\left( {\ln x} \right)}^{ - \ln x}}\cdot1} \right)} \;dx\\
 = {\left( {\ln x} \right)^{ - \ln x}}\int 1 \;dx - \int {\left( { - \frac{{{{\left( {\ln x} \right)}^{ - \ln x}}}}{x}\;\cdot\int 1 \;dx} \right)} \;dx\\
 = x{\left( {\ln x} \right)^{ - \ln x}} - \int {\left( { - {{\left( {\ln x} \right)}^{ - \ln x}}} \right)} \;dx\\
 = x{\left( {\ln x} \right)^{ - \ln x}} + {\left( {\ln x} \right)^{ - \ln x}}\int 1 \;dx + \int {\left( {\frac{{{{\left( {\ln x} \right)}^{ - \ln x}}}}{x}\;\int 1 \;dx} \right)\;dx} \\
 = x{\left( {\ln x} \right)^{ - \ln x}} + x{\left( {\ln x} \right)^{ - \ln x}} + {\left( {\ln x} \right)^{ - \ln x}}\int 1 \;dx - \int {\left( {\frac{{ - {{\left( {\ln x} \right)}^{ - \ln x}}}}{x}\int 1 \;dx} \right)} \;dx\\
 = \sum\limits_{n = 1}^\infty  {\left( {x{{\left( {\ln x} \right)}^{ - \ln x}}} \right)} \\
so\\
\int_1^n {{{\left( {\ln x} \right)}^{ - \ln x}}} \;dx = \sum\limits_{k = 1}^\infty  {\left( {n{{\left( {\ln n} \right)}^{ - \ln n}}} \right)} \; - \sum\limits_{k = 1}^\infty  {\left( {{{\left( {\ln 1} \right)}^{ - \ln 1}}} \right)}  = \sum\limits_{k = 1}^\infty  {\left( {n{{\left( {\ln n} \right)}^{ - \ln n}}} \right)}  - \sum\limits_{k = 1}^\infty  {\left( {{0^0}} \right)} 
\end{array}
