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? 
 A: Bad idea. Most likely, the integrand will not have elementary primitive. Try another convergence test.
A: By exponentiating the logarithm, we observe that $(\ln x)^{\ln x} = x^{\ln \ln x}$. Now substitute $u = e^x$ to obtain
$$\int_0^\infty \frac{e^{u}}{e^{u \log u}} \; du$$
which is finite by comparison to $\int_0^\infty \frac{1}{e^u} \; du$
A: This integral likely can't be evaluated exactly, but we can approximate it for sufficiently large $n$.  For $n > e^{e^2}$, we have that
$$\int\limits_1^{n} \frac{1}{\ln(x)^{\ln(x)}} dx = \int\limits_1^{e^{e^2}} \frac{1}{\ln(x)^{\ln(x)}} dx + \int\limits_{e^{e^2}}^{n} \frac{1}{\ln(x)^{\ln(x)}} dx.$$
Because the first integral on the RHS 'plays nice' (that is, does not blow up to infinity) it has a finite value, which we will henceforth call $I$.  Based on WolframAlpha, $I \approx 6.39951$.  To evaluate the second integral, we can perform the substitution $u = \ln(x)$.  This yields $du = \frac{1}{x}dx$, so $e^u du = dx$.  Thus, the original integral can be written as
$$I + \int\limits_{e^2}^{\ln(n)} \frac{e^u}{u^{u}} du = I + \int\limits_{e^2}^{\ln(n)} \left(\frac{e}{u}\right)^u du.$$
Over the region of integration, the integrand of the second integral is less  than or equal to $e^{-u}$, so we have that 
$$\int\limits_1^{n} \frac{1}{\ln(x)^{\ln(x)}} dx \le I + \int\limits_{e^2}^{\ln(n)} e^{-u} du = I + e^{-e^2} -\frac{1}{n}.$$
Taking the limit as $n \to \infty$, we have that 
$$\int\limits_1^{n} \frac{1}{\ln(x)^{\ln(x)}} dx \le I + e^{-e^2} \approx 6.4$$
This agrees with the WolframAlpha computation of the integral for $n=10000$.
A: \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}
