How to solve the following integral with natural logs:
$$\int_0^e \frac 1{\ln (x^x)} dx$$
Can anyone help me to solve this.
This is an improper integral which diverges. $$\int_0^e \frac 1{\ln (x^x)} dx = \int_0^e \frac 1{x\ln (x)} dx $$
Upon substitution of $$ u=\ln x$$
We get $$ \int_{-\infty}^1\frac {1}{u} du =$$
$$\ln (|u| )|_{-\infty}^1 = -\infty$$
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This is an improper integral which diverges. $$\int_0^e \frac 1{\ln (x^x)} dx = \int_0^e \frac 1{x\ln (x)} dx $$
Upon substitution of $$ u=\ln x$$
We get $$ \int_{-\infty}^1\frac {1}{u} du =$$
$$\ln (|u| )|_{-\infty}^1 = -\infty$$