That doesn't look integrable

Calculus Level 4

$\sum_{n=1}^\infty\int_{\frac{1}{e^n}}^0\left(\log_xe-(\log_xe)^2\right)\text{ }dx=A-\log_B(C-D)$

Given that $A$ and $D$ are positive integers, find $A+\dfrac{B}{C}+D$ .

The answer is 3.

1 solution

Karan Shekhawat
Apr 22, 2015

First Note This is integration of type : $\int { \left( f\left( x \right) +xf^{ ' }\left( x \right) \right) dx } =xf\left( x \right) +c\\ f(x)=\cfrac { 1 }{ \ln { x } }$

Now put limits and now our task is to find : $g(1/e)=?$ where : $\displaystyle{g(x)=\sum _{ 1 }^{ n }{ \cfrac { x^{ n } }{ n } } \\ g^{ ' }\left( x \right) =\sum _{ 1 }^{ n }{ x^{ n-1 } } =1+x+{ x }^{ 2 }+.....\\ g^{ ' }\left( x \right) =\cfrac { 1 }{ 1-x } \\ g(x)=-\ln { (1-x) } \\ \boxed { g(\cfrac { 1 }{ e } )=1-\ln { (e-1) } } }$

That doesn't look integrable

The answer is 3.

1 solution

0 pending reports