Ramanujan is a true master!

Calculus Level 5

If, ∫ 0 ∞ x 1726 + x 1725 + x 1724 + . . . + x 2 + x + 1 x 1728 + x 1727 + x 1726 + . . . + x 2 + x + 1 d x = A π B cot ⁡ ( π C ) \int _{ 0 }^{ \infty }{ \frac { { x }^{ 1726 }+{ x }^{ 1725 }+{ x }^{ 1724 }+...+{ x }^{ 2 }+{ x }+1 }{ { x }^{ 1728 }+{ x }^{ 1727 }+{ x }^{ 1726 }+...+{ x }^{ 2 }+{ x }+1 } dx } =\frac { A\pi }{ B } \cot\left( \frac { \pi }{ C } \right)

A , B , C > 0 A,B,C>0 and g c d ( A , B ) = 1 gcd(A,B)=1

Find: A + B + C A+B+C

For more calculus problems see this.

75% original


The answer is 3460.

Aditya Kumar by Aditya Kumar , 22, India

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1 solution

Aditya Kumar
Oct 16, 2015

I = ∫ 0 ∞ x 1726 + x 1725 + x 1724 + . . . + x 2 + x + 1 x 1728 + x 1727 + x 1726 + . . . + x 2 + x + 1 d x = ∫ 0 ∞ 1 − x 1727 1 − x 1729 d x L e t , I 1 = ∫ 0 ∞ 1 1 − x 1729 d x a n d I 2 = ∫ 0 ∞ x 1727 1 − x 1729 d x N o w , s o l v i n g I 1 L e t x 1729 = t ⇒ d x = 1 1729 t − 1728 1729 I 1 = 1 1729 ∫ 0 ∞ t − 1728 1729 1 − t d t 1 1729 { M ( 1 1 − x ) } ( 1 1729 ) { M f } ( s ) ⇒ M e l l i n ′ s t r a n s o r m o f f a t s N o w , u s i n g R a m a n u j a n ′ s m a s t e r t h e o r e m , I 1 = 1 1729 Γ ( 1 1729 ) Γ ( 1 − 1 1729 ) ( e i π 1729 ) U s i n g E u l e r ′ s R e f l e c t i o n F o r m u l a , I 1 = 1 1729 π s i n ( π 1729 ) ( e i π 1729 ) N o w , s o l v i n g I 2 L e t x 1729 = t ⇒ d x = 1 1729 t − 1728 1729 I 2 = 1 1729 ∫ 0 ∞ x − 1 1729 1 − x d x ) = 1 1729 M ( 1 1 − x ) ( 1728 1729 ) ) S o l v i n g a s b e f o r e , I 2 = 1 1729 π s i n ( 1728 π 1729 ) ( e i π 1729 ) P u t t i n g t h e m i n o r i g i n a l e q n , I = 1 1729 ( π s i n ( π 1729 ) ) ( e i π 1729 + e − i π 1729 ) ∴ I = 2 π 1729 c o t ( π 1729 ) I=\int _{ 0 }^{ \infty }{ \frac { { x }^{ 1726 }+{ x }^{ 1725 }+{ x }^{ 1724 }+...+{ x }^{ 2 }+{ x }+1 }{ { x }^{ 1728 }+{ x }^{ 1727 }+{ x }^{ 1726 }+...+{ x }^{ 2 }+{ x }+1 } dx } \\ \quad =\int _{ 0 }^{ \infty }{ \frac { 1-{ x }^{ 1727 } }{ 1-{ x }^{ 1729 } } } dx\\ Let,\\ { I }_{ 1 }=\int _{ 0 }^{ \infty }{ \frac { 1 }{ 1-{ x }^{ 1729 } } dx } \quad and\quad { I }_{ 2 }=\int _{ 0 }^{ \infty }{ \frac { { x }^{ 1727 } }{ 1-{ x }^{ 1729 } } dx } \\ Now,\quad solving\quad { I }_{ 1 }\\ Let\quad \\ { x }^{ 1729 }=t\quad \Rightarrow dx=\frac { 1 }{ 1729 } { t }^{ \frac { -1728 }{ 1729 } }\\ { I }_{ 1 }=\frac { 1 }{ 1729 } \int _{ 0 }^{ \infty }{ \frac { { t }^{ \frac { -1728 }{ 1729 } } }{ 1-t } dt } \\ \frac { 1 }{ 1729 } \left\{ M\left( \frac { 1 }{ 1-x } \right) \right\} \left( \frac { 1 }{ 1729 } \right) \\ \left\{ Mf \right\} (s)\Rightarrow Mellin's\quad transorm\quad of\quad f\quad at\quad s\\ Now,\quad using\quad Ramanujan's\quad master\quad theorem,\\ { I }_{ 1 }=\frac { 1 }{ 1729 } \Gamma \left( \frac { 1 }{ 1729 } \right) \Gamma \left( 1-\frac { 1 }{ 1729 } \right) \left( { e }^{ \frac { i\pi }{ 1729 } } \right) \\ UsingEuler'sReflectionFormula,\\ { I }_{ 1 }=\frac { 1 }{ 1729 } \frac { \pi }{ sin\left( \frac { \pi }{ 1729 } \right) } \left( { e }^{ \frac { i\pi }{ 1729 } } \right) \\ Now,\quad solving\quad { I }_{ 2 }\\ Let\quad \\ { x }^{ 1729 }=t\quad \Rightarrow dx=\frac { 1 }{ 1729 } { t }^{ \frac { -1728 }{ 1729 } }\\ \\ { I }_{ 2 }=\frac { 1 }{ 1729 } \int _{ 0 }^{ \infty }{ \frac { { x }^{ \frac { -1 }{ 1729 } } }{ 1-x } dx } )\\ \quad =\frac { 1 }{ 1729 } M\left( \frac { 1 }{ 1-x } \right) \left( \frac { 1728 }{ 1729 } \right) )\\ Solving\quad as\quad before,\\ { I }_{ 2 }=\frac { 1 }{ 1729 } \frac { \pi }{ sin\left( \frac { 1728\pi }{ 1729 } \right) } \left( { e }^{ \frac { i\pi }{ 1729 } } \right) \\ Putting\quad them\quad in\quad original\quad eqn,\\ I=\frac { 1 }{ 1729 } \left( \frac { \pi }{ sin\left( \frac { \pi }{ 1729 } \right) } \right) \left( { e }^{ \frac { i\pi }{ 1729 } }+{ e }^{ -\frac { i\pi }{ 1729 } } \right) \\ \therefore \quad I=\frac { 2\pi }{ 1729 } cot\left( \frac { \pi }{ 1729 } \right) \\

4 Helpful 0 Interesting 0 Brilliant 0 Confused

As far as I can remember, this must be inspired by some Pi Han Goh's problem for which I have written quite the same solution, right?

BTW, nice solution! It is for the first time so far that I am seeing RMT in someone else's solution.

Kartik Sharma - 5 years, 8 months ago

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Yes. Others also know rmt. I asked Parth. He also knows rmt. I've learnt rmt. I'm using it in various problems.

Aditya Kumar - 5 years, 8 months ago

@Kartik Sharma can u post problems which use rmt. I need a bit of practice. Or can u suggest me a book from where I could practice problems involving rmt. This is a humble request. :)

Aditya Kumar - 5 years, 8 months ago

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Well, I have not seen RMT in any full course calculus book yet. But there are some specific texts written on this topic and which are worth to be read. You can see this .

Kartik Sharma - 5 years, 7 months ago

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@Kartik Sharma – Thanks for the PDF! I'll surely work on it!

Aditya Kumar - 5 years, 7 months ago

@Abhishek Bakshi try this then comment.

Aditya Kumar - 5 years, 8 months ago

@Calvin Lin sir can u edit the last line of the solution to ∴ I = 2 π 1729 c o t ( π 1729 ) \therefore \quad I=\frac { 2\pi }{ 1729 } cot\left( \frac { \pi }{ 1729 } \right) \\ . I'm sorry :'( for not being able to edit it.

Aditya Kumar - 5 years, 7 months ago

Note that you need to justify one step in between where you multiplied the numerator and denominator by 1 − x 1-x which would create problems when x = 1 x=1 since the limits of the integration are 0 0 to ∞ \infty .

Rohan Shinde - 1 year, 6 months ago

Else otherwise you can split the limits from 0 0 to 1 1 and 1 1 to ∞ \infty and substitute x = 1 / t x=1/t in the second part and then use integral definition of Digamma Function.

Rohan Shinde - 1 year, 6 months ago

I don't get what's going on here. The integrals I1 and I2 diverge.

Joe Mansley - 4 months, 3 weeks ago

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