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Calculus Level 5

Let ϕ ( x ) = n = 1 ( 1 x n ) \phi\left(x\right)=\prod_{n=1}^{\infty}\left(1-x^n\right)

If

0 1 ϕ ( x ) d x = a b c π sinh ( c π b d ) d cosh ( c π b ) f \int_0^1\phi\left(x\right)dx=\frac{a\sqrt{\frac{b}{c}}\pi\sinh\left(\frac{\sqrt{c}\pi}{b\cdot d}\right)}{d\cosh\left(\frac{\sqrt{c}\pi}{b}\right)-f}

where a , b , c , d , f \displaystyle a,b,c,d,f are positive integers, then find the value of a + b + c + d + f \displaystyle a+b+c+d+f


The answer is 37.

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1 solution

Syed Shahabudeen
May 29, 2020

\(\begin{align*} \prod _{ n=0 }^{ \infty }{ \left( 1-{ x }^{ n+1 } \right) =\prod _{ n=0 }^{ \infty }{ \left( 1-{ x }^{ 3n+1 } \right) \left( 1-{ x }^{ 3n+2 } \right) \left( 1-{ x }^{ 3n+3 } \right) } } &&&&&&&&&&&& \textcolor{green}{\textrm{Jacobi triple product}\sum _{ n=-\infty }^{ \infty }{ { q }^{ { n }^{ 2 } }{ z }^{ n }=\prod _{ n=0 }^{ \infty }{ \left( 1-{ q }^{ 2n+2 } \right) \left( 1+{ q }^{ 2n+1 }z \right) \left( 1+{ q }^{ 2n+1 }{ z }^{ -1 } \right) } } ;\quad q={ x }^{ \frac { 3 }{ 2 } }, z=-{ x }^{ \frac { -1 }{ 2 } }}\\

\end{align*}\) therefore we get ϕ ( x ) = n = 0 ( 1 x 3 n + 1 ) ( 1 x 3 n + 2 ) ( 1 x 3 n + 3 ) = n = ( 1 ) n x n ( 3 n 1 ) 2 \begin{aligned}\quad \phi \left( x \right) ={\prod _{ n=0 }^{ \infty }{ \left( 1-{ x }^{ 3n+1 } \right) \left( 1-{ x }^{ 3n+2 } \right) \left( 1-{ x }^{ 3n+3 } \right) } }= &\sum _{ n=-\infty }^{ \infty }{ { \left( -1 \right) }^{ n }{ x }^{ \frac { n\left( 3n-1 \right) }{ 2 } } }\end{aligned} 0 1 ϕ ( x ) = 0 1 n = ( 1 ) n x n ( 3 n 1 ) 2 = n = 2 ( 1 ) n 3 n 2 n + 2 n = ( 1 ) n a n 2 + b n + c = 1 b 2 4 a c 4 π sin ( π b 2 4 a c 2 a ) cos ( π b 2 a ) cos ( π b 2 4 a c a ) cos ( π b a ) = 8 3 23 π sinh ( π 23 6 ) 2 cosh ( π 23 3 ) 1 \begin{aligned} \int _{ 0 }^{ 1 }{ \phi \left( x \right) } &=\int _{ 0 }^{ 1 }{ \sum _{ n=-\infty }^{ \infty }{ { \left( -1 \right) }^{ n }{ x }^{ \frac { n\left( 3n-1 \right) }{ 2 } } } }\\& =\sum _{ n=-\infty }^{ \infty }{ \frac { 2{ \left( -1 \right) }^{ n } }{ 3{ n }^{ 2 }-n+2 } }&&&&&&&&&&&&&&&&\textcolor{#BA33D6}{ \sum _{ n=-\infty }^{ \infty }{ \frac { { \left( -1 \right) }^{ n } }{ a{ n }^{ 2 }+bn+c } } =\frac { 1 }{ \sqrt { { b }^{ 2 }-4ac } } \frac { 4\pi \sin \left( \frac { \pi \sqrt { { b }^{ 2 }-4ac } }{ 2a } \right) \cos \left( \frac { \pi b }{ 2a } \right) }{ \cos \left( \frac { \pi \sqrt { { b }^{ 2 }-4ac } }{ a } \right) -\cos \left( \frac { \pi b }{ a } \right) }}\\&=\frac { 8\sqrt { \frac { 3 }{ 23 } } \pi \sinh \left( \frac { \pi \sqrt { 23 } }{ 6 } \right) }{ 2\cosh \left( \frac { \pi \sqrt { 23 } }{ 3 } \right) -1 } \end{aligned} so we get a = 8 , b = 3 , c = 23 , d = 2 , f = 1 a=8,b=3,c=23,d=2,f=1 a + b + c + d + f = 37 \boxed{a+b+c+d+f=37}

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