Another Logarithm Problem.(14)

$\large \displaystyle \text{By chain rule, } \frac{\text{d}}{\text{d}x} \left(\ln x \right)^{n} = \frac{n \ln^{n-1} x}{x}$

$\large \displaystyle \text{By using integration By Parts}$

$\large \displaystyle \int_{1}^{\infty} \frac{(\ln x)^{2011}}{x^{2011}} \, dx = \left[\frac{(\ln x)^{2011}}{-2010 x^{2010}} \right]_1^{\infty} - \int_1^{\infty} \frac{n(\ln x)^{n-1}}{-2010x^{2011}} \, dx$

$\large \displaystyle \ln(1) = 0, \lim_{x \rightarrow \infty} \frac{(\ln x)^{n}}{x^{2010}} = 0 \, \forall \, n > 0.$

$\large \displaystyle \int_1^{\infty} \frac{(\ln x)^{2011}}{x^{2011}} \, dx = \int_1^{\infty} \frac{n (\ln x)^{n-1}}{2010 x^{2011}} \, dx$

$\large \displaystyle \implies \int_1^{\infty} \frac{(\ln x)^{2011}}{x^{2011}} \, dx = \frac{2011!}{2010^{2011}} \int_1^{\infty} \frac{1}{x^{2011}} \, dx = \frac{\color{#D61F06}{2011!}}{\color{#20A900}{2010}^{\color{#3D99F6}{2012}}}$ .

$\huge \displaystyle \frac{\color{#D61F06}{b!}}{\color{#20A900}{a}^{\color{#3D99F6}{c}}} = \frac{\color{#D61F06}{2011!}}{\color{#20A900}{2010}^{\color{#3D99F6}{2012}}}$

$\huge \displaystyle \color{#20A900}{a} + \color{#D61F06}{b} + \color{#3D99F6}{c} = \color{#20A900}{2010} + \color{#D61F06}{2011} + \color{#3D99F6}{2012} = \color{#EC7300}{\boxed{6033}}$