Floors and Ceilings 2 - Room on Ceiling, Floor on Room

$\displaystyle\int\limits_{0}^{\infty} \frac{x}{\lceil x \rceil} - \frac{\lfloor x \rfloor}{x} dx$

Let $\displaystyle I_1 = \int\limits_{0}^{\infty} {\frac{x}{\lceil x \rceil}} dx$ and $\displaystyle I_2 = \int\limits_0^{\infty} {\frac{\lfloor x \rfloor}{x}} dx$

Breaking at the integral points $I_1 = \displaystyle\lim_{n \to \infty} [\int\limits_0^1 {\frac{x}{1}} dx + \int\limits_1^2 {\frac{x}{2}} dx + \int\limits_2^3 {\frac{x}{3}} dx + ......... + \int\limits_{n-1}^{n} {\frac{x}{n}} dx]$ $=\displaystyle\lim_{n \to \infty} [\sum_{r=1}^{n} (1 - \frac{1}{2r})]$ $=\displaystyle\lim_{n \to \infty} (n - H_n/2)$

The series expansion of the above expression is :

$\displaystyle I_1 = n + \frac{1}{2}\log(\frac{1}{2}) - \frac{1}{2}\gamma - \frac{1}{4n} + \frac{1}{240n^4} + . . . . .$

SIMILARLY , Breaking at the integral points for the other integral $I_2$ , we get

$I_2 = \displaystyle\lim_{n \to \infty} [\int\limits_0^1 {\frac{0}{x}} dx + \int\limits_1^2 {\frac{1}{x}} dx + \int\limits_2^3 {\frac{2}{x}} dx + ........ + \int\limits_{n-1}^{n} {\frac{n-1}{x}} dx$ $=\displaystyle\lim_{n \to \infty}[\sum_{r=1}^{n-1} r\log(\frac{r+1}{r})]$

The series expansion of the above expression is :

$\displaystyle I_2 = n(-\log(\frac{1}{n}) - \log(n) + 1) + \frac{1}{2}\log(\frac{1}{2n\pi}) - \frac{1}{12n} + \frac{1}{360n^3} + . . . . . . . .$ $\displaystyle I_2 = n + \frac{1}{2}\log(\frac{1}{2n\pi}) - \frac{1}{12n} + \frac{1}{360n^3} + . . . . . . . .$

Therefore, $\displaystyle I_1 - I_2 = n - n + \frac{1}{2} (\log(\frac{1}{n}) - \log(\frac{1}{2n\pi})) - \frac{1}{2}\gamma - \frac{1}{6n} + .......$ (other terms which will coverge to zero as n tends to infinity ) $\displaystyle I_1 - I_2 = \frac{1}{2}\log(2\pi) - \frac{1}{2}\gamma + 0 + 0 + . . . . .$

Hence, $\displaystyle p + 2q + 3r + 4s + 5t = 1 + 4 + 3 + 8 + 5 = 21$

The required answer is $\displaystyle 21$ .

Thanks for reading the solution . :)

Nice problem!

$\displaystyle \int \limits_{0}^{\infty}{\left(\frac{x}{\lceil x \rceil} - \frac{\lfloor x \rfloor}{x} \right)} dx$

We can write it as -

$\displaystyle \sum_{k=0}^{\infty}{\left(\int \limits_{k}^{k+1}{\frac{x}{\lceil x \rceil}} - \int \limits_{k}^{k+1}{\frac{\lfloor x \rfloor}{x}}\right)}$

$\displaystyle \sum_{k=0}^{\infty}{\left(\frac{1}{k+1}\int \limits_{k}^{k+1}{x} - k\int \limits_{k}^{k+1}{\frac{1}{x}}\right)}$

Let's straightaway do the integral and convert it into a sum,

$\displaystyle \sum_{k=0}^{\infty}{\frac{2k+1}{2k+2}} - \sum_{k=0}^{\infty}{k ln(k+1)} + \sum_{k=0}^{\infty}{k lnk}$

$\displaystyle \sum_{k=0}^{\infty}{1} - \sum_{k=1}^{\infty}{\frac{1}{2k}} - \sum_{k=1 }^{\infty}{k lnk} + \sum_{k=1}^{\infty}{ln(k)} + \sum_{k=0}^{\infty}{k lnk}$

Now, we must consider one series which is famously known as Zeta function -

$\displaystyle \zeta(s) = \sum_{n=1}^{\infty}{\frac{1}{{n}^{s}}}$

Differentiate both sides w.r.t. s, we get

$\displaystyle \zeta'(s) = \sum_{n=1}^{\infty}{-\frac{ln(n)}{{n}^{s}}}$

We want one value i.e. $\displaystyle \zeta'(0) = \frac{-1}{2} ln(2\pi)$

Hence, we have got the value of the sum $\displaystyle \sum_{k=1}^{\infty}{ln(k)}$

We can write our sum

$\displaystyle \frac{1}{2} ln(2\pi) - \frac{1}{2}(\lim_{n->\infty}{\sum_{k=1}^{n}{\frac{1}{k}} - ln(n)})$

which is just

$\displaystyle \frac{1}{2} ln(2\pi) - \frac{1}{2}\gamma$

$Sorry\quad for\quad my\quad lengthy\quad proof.$

$Our\quad integral\quad can\quad be\quad written\quad as\quad \sum _{ n=0 }^{ \infty }{ \int _{ n }^{ n+1 }{ \left( \frac { x }{ \left\lceil x \right\rceil } -\frac { \left\lfloor x \right\rfloor }{ x } \right) . } } \\ \\ This\quad can\quad again\quad be\quad written\quad as\quad \sum _{ n=0 }^{ \infty }{ \int _{ n }^{ n+1 }{ \left( \frac { x }{ n+1 } -\frac { n }{ x } \right) \quad } } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\sum _{ n=0 }^{ \infty }{ \left( \frac { 1 }{ 2 } \frac { 2n+1 }{ n+1 } -n\ln { \left( \frac { n+1 }{ n } \right) } \right) } \\ \\ We're\quad gonna\quad rewrite\quad the\quad above\quad summation\quad as\quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \underset { m\rightarrow \infty }{ lim } \sum _{ n=0 }^{ m }{ \left( \frac { 1 }{ 2 } \frac { 2n+1 }{ n+1 } -n\ln { \left( \frac { n+1 }{ n } \right) } \right) } \\ \quad \quad \quad \quad \quad \quad \quad =\underset { m\rightarrow \infty }{ lim } \sum _{ n=0 }^{ m }{ \left( 1-\frac { 1 }{ 2n+2 } -\left( n\ln { (n+1 } \right) -n\ln { (n) } ) \right) } \\ \quad \quad \quad \quad \quad =\underset { m\rightarrow \infty }{ lim } \left( (m+1)-(m\ln { (m+1) } +\ln { (m! } \right) -\frac { 1 }{ 2 } \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ n+1 } } )\\ I've\quad not\quad stated\quad how\quad to\quad sum\quad n\ln { (n+1) } -n\ln { (n) } \\ as\quad it\quad should\quad be\quad easy\quad for\quad a\quad Level\quad 4\quad or\quad 5\quad student.\\ \\ =\underset { m\rightarrow \infty }{ lim } \left( (m+1)-m\ln { (m+1) } +\ln { (m! } \right) -\frac { 1 }{ 2 } \ln { (m+1) } -(\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ n } } -\frac { 1 }{ 2 } \ln { (m+1) } ))\\ =\lim _{ m\rightarrow \infty }{ \ln { \frac { { e }^{ m+1 }m! }{ { m+1 }^{ m+0.5 } } } } -\frac { \gamma }{ 2 }$

$Now$ $using$ Stirling's Formula $,$ $for$ $very$ $large$ $values$ $of$ $m,$ $m!$ $tends$ $to$ $\sqrt { 2\pi m } { \left( \frac { m }{ e } \right) }^{ m }.$

$Using\quad this\quad in\quad our\quad limit,\quad we\quad get\quad it\quad as\quad \\ \lim _{ m\rightarrow \infty }{ \ln { \frac { { e }^{ m+1 }\sqrt { 2\pi m } { \left( \frac { m }{ e } \right) }^{ m } }{ { \left( m+1 \right) }^{ \left( m+0.5 \right) } } } } -\frac { \gamma }{ 2 } \\ As\quad the\quad logarithm\quad function\quad is\quad continuous\quad our\quad limit\quad can\quad be\quad \\ written\quad as\quad \ln { \left( \lim _{ m\rightarrow \infty }{ \frac { { e }^{ m+1 }\sqrt { 2\pi m } { \left( \frac { m }{ e } \right) }^{ m } }{ { \left( m+1 \right) }^{ \left( m+0.5 \right) } } } \right) } -\frac { \gamma }{ 2 } \\ \\ =\ln { \left( \lim _{ m\rightarrow \infty }{ e\sqrt { 2\pi } { \left( \frac { m }{ m+1 } \right) }^{ m+0.5 } } \right) } -\frac { \gamma }{ 2 } \quad \\ =\ln { \left( \lim _{ m\rightarrow \infty }{ e\sqrt { 2\pi } { \left( 1-\frac { 1 }{ m+1 } \right) }^{ m+0.5 } } \right) } -\frac { \gamma }{ 2 } \\ \\ =\ln { \left( \lim _{ m\rightarrow \infty }{ e\sqrt { 2\pi } { \left( 1-\frac { 1 }{ m+1 } \right) }^{ -\frac { m+0.5 }{ m+1 } } } \right) } -\frac { \gamma }{ 2 } .\quad \\ Computing\quad this\quad limit,\quad we\quad get\quad \\ \\ \ln { \left( e\sqrt { 2\pi } { e }^{ -1 } \right) } -\quad \frac { \gamma }{ 2 } =\quad \boxed { \frac { 1 }{ 2 } \ln { (2\pi )-\frac { \gamma }{ 2 } } } .\\ \\ Hence\quad p=1,q=2,r=1,s=2,t=1\quad \\ and\quad p+2q+3r+4s+5t=\boxed { 21 }$

$Phew!\quad Thanks\quad for\quad reading\quad the\quad full\quad solution.$