A nice integral

Let $\displaystyle \ln x = t$
Then,
$\displaystyle dx = e^t dt$

The integral simplifies to,

$\displaystyle \int\limits_0^{\infty} t^{2014} e^{-2013t} dt$

Using IBP, we get,

$\displaystyle \int\limits_0^{\infty} t^{2014} e^{-2013t} dt = \left.\frac{t^{2014}e^{-2013t}}{-2013}\right|_0^{\infty} + \frac{2014}{2013}\int\limits_0^{\infty} t^{2013}e^{-2013t}dt$

$\displaystyle I = 0 + \frac{2014}{2013}\int\limits_0^{\infty} t^{2013}e^{-2013t}dt$

Repeating the same procedure, we get,

$\displaystyle I = \frac{2014}{2013}(0+\frac{2013}{2013}\int\limits_0^{\infty} t^{2012} e^{-2013t} dt)$

Proceeding this way till the power of $\displaystyle t$ becomes $\displaystyle 0$ , we get,

$\displaystyle I = \frac{2014}{2013}\times\frac{2013}{2013}\times\ldots\times\frac{1}{2013}\int\limits_0^{\infty} e^{-2013t} dt$

Thus,
$\displaystyle I = \frac{2014!}{2013^{2014}} \times \frac{1}{2013} = \frac{2014!}{2013^{2015}} = \frac{p!}{q^r}$

Hence,
$\displaystyle p+q+r = 2014+2013+2015 = \boxed{6042}$

We write $f(\alpha )=\int _{ 1 }^{ \infty }{ { x }^{ \alpha }dx } =\frac { -1 }{ \alpha +1 } \quad for\quad \alpha <-1$

Differentiating both sides with respect to $\alpha$

2014 times we get :

$\frac { -(2014)! }{ { (\alpha +1) }^{ 2015 } } =\int _{ 1 }^{ \infty }{ { ln }^{ 2014 }(x){ x }^{ \alpha }dx }$

Simply put $\alpha =-2014$ to get :

$\boxed{\int _{ 1 }^{ \infty }{ { (\frac { ln(x) }{ x } })^{ 2014 }dx } =\frac { (2014)! }{ { 2013 }^{ 2015 } }}$

2014!/ 2013^2015 = (3.2554396263186602540430307685016+) x 10^(-876)

We must apply a genuine technique for this! S (u d v) = u v - S (v d u) of zero u v with 0 to infinity:

Converted into definite integral of [e^(2013 y)] (y^2014) d y from 0 to infinity with y = Ln x,

With integral of [e^(2013 y)] (y^n) d y = I (n)

I (2014) = (2014/ 2013)(2013/ 2013)(2012/ 2013) I (2011) = 2014!/ 2013^2014 I (0)

I (0) = Integral of e^(2013 y) d y from 0 to infinity = 1/ 2013

I (2014) = 2014!/ 2013^2015

2014 + 2013 + 2015 = 6042

HMMT 2011 Problem 11.