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Calculus Level 5

∫ 0 ∞ d x x 2 / 3 ( 1 + 2 x + x 2 ) \displaystyle \large\int_{0}^{\infty}{\frac{dx}{{x}^{2/3}(1+\sqrt{2}x+{x}^{2})}}

If the integral above evaluates to π a b c \displaystyle \frac{{\pi}^{a}\sqrt{b}}{c} for positive integers a , b , c a,b,c with b b square-free, then submit your answer as a + b + c a+b+c .


The answer is 10.

Kartik Sharma by Kartik Sharma , 21, India

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2 solutions

Ronak Agarwal
Dec 29, 2015

Why even use Ramanujan Master's Theorom :

Check this :

The integral is equal to : I = ∫ 0 ∞ d x x 2 / 3 ( x 2 + 2 x + 1 ) = ∫ 0 ∞ ( x 2 − 2 x + 1 ) d x x 2 / 3 ( x 4 + 1 ) \displaystyle I=\int _{ 0 }^{ \infty }{ \frac { dx }{ { x }^{ 2/3 }({ x }^{ 2 }+\sqrt { 2 } x+1) } } =\int _{ 0 }^{ \infty }{ \frac { ({ x }^{ 2 }-\sqrt { 2 } x+1)dx }{ { x }^{ 2/3 }({ x }^{ 4 }+1) } }

Use the result :

∫ 0 ∞ x m − 1 1 + x n d x = π n csc ⁡ ( m π n ) \displaystyle\int _{ 0 }^{ \infty }{ \frac { { x }^{ m-1 } }{ 1+{ x }^{ n } } dx } =\frac { \pi }{ n } \csc { \left( \frac { m\pi }{ n } \right) }

To get the answer as

I = π 4 csc ⁡ ( 7 π 12 ) − 2 π 4 csc ⁡ ( π 3 ) + π 4 csc ⁡ ( π 12 ) = π 6 3 \displaystyle I=\frac { \pi }{ 4 } \csc { \left( \frac { 7\pi }{ 12 } \right) } -\sqrt { 2 } \frac { \pi }{ 4 } \csc { \left( \frac { \pi }{ 3 } \right) } +\frac { \pi }{ 4 } \csc { \left( \frac { \pi }{ 12 } \right) } =\frac { \pi \sqrt { 6 } }{ 3 }

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Same Here :)

Richeek Das - 2 years, 5 months ago
Kartik Sharma
Aug 8, 2015

I don't know an alternate solution and I care not for it. I have got something which is too exciting(at least for me).

And drum rollers beating, it's Ramanujan Master Theorem .

Chebyshev polynomials of the 2nd kind -

U n ( a ) = s i n ( ( n + 1 ) x ) s i n ( x ) , a = c o s ( x ) {U}_{n}(a) = \frac{sin((n+1)x)}{sin(x)}, a = cos(x)

Then, the generating function

∑ k = 1 ∞ U k ( a ) x k = 1 1 − 2 a + x 2 \displaystyle \sum_{k=1}^{\infty}{{U}_{k}(a){x}^{k}} = \frac{1}{1-2a+{x}^{2}} [I've provided just the results, you can derive them or check them out at Chebyshev wiki]

Now, according to Ramanujan Master Theorem,

∫ 0 ∞ x s − 1 d x 1 + 2 a + x 2 = π s i n ( π s ) s i n ( ( 1 − s ) c o s − 1 ( a ) ) 1 − a 2 \displaystyle \int_{0}^{\infty}{\frac{{x}^{s-1} dx}{1+2a+{x}^{2}}} = \frac{\pi}{sin(\pi s)}\frac{sin((1-s){cos}^{-1}(a))}{\sqrt{1-{a}^{2}}}

This has come so by usual application of the Theorem - using Reflection formula, transformation into hypergeometric function and so on.

Substitute s = 1 3 s = \frac{1}{3} and you are done.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Rightly called the MASTER Theorem! I will post more.

Kartik Sharma - 5 years, 10 months ago

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This is probably the best solution in which I understood none of it. Still upvoted!

Pi Han Goh - 5 years, 10 months ago

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Lol. Try to read the wiki I've linked. You will surely find it helpful.

Kartik Sharma - 5 years, 10 months ago

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@Kartik Sharma –

Now, according to Ramanujan Master Theorem,

∫ 0 ∞ x s − 1 d x 1 + 2 a + x 2 = π s i n ( π s ) s i n ( ( 1 − s ) c o s − 1 ( a ) ) 1 − a 2 \displaystyle \int_{0}^{\infty}{\frac{{x}^{s-1} dx}{1+2a+{x}^{2}}} = \frac{\pi}{sin(\pi s)}\frac{sin((1-s){cos}^{-1}(a))}{\sqrt{1-{a}^{2}}}

Where did you get this? I briefly skimmed through the link you provided but I don't see any expression closely related to this.

Pi Han Goh - 5 years, 10 months ago

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@Pi Han Goh – You must have then calculated the generating function involving Chebyshev polynomials? I think that it then becomes obvious that we then use Chebyshev polynomials.

M ( 1 1 + 2 a + x 2 ) = Γ ( s ) ϕ ( − s ) \displaystyle M\left(\frac{1}{1+2a+{x}^{2}}\right) = \Gamma(s)\phi(-s)

where ϕ ( k ) \displaystyle \phi(k) is the coefficient of ( − x ) k k ! \displaystyle \frac{{(-x)}^{k}}{k!} in the generating function of 1 1 + 2 a + x 2 \displaystyle \frac{1}{1+2a+{x}^{2}} and M ( f ) M(f) is the Mellin transform of f f .

We know ∑ k = 1 ∞ U k ( − a ) ( − x ) k = 1 1 + 2 a + x 2 \displaystyle \sum_{k=1}^{\infty}{{U}_{k}(-a){(-x)}^{k}} = \frac{1}{1+2a+{x}^{2}}

so that ϕ ( k ) = Γ ( k + 1 ) U k ( − a ) \displaystyle \phi(k) = \Gamma(k+1){U}_{k}(-a)

Using Ramanujan Master Theorem now,

M ( 1 1 + 2 a + x 2 ) = Γ ( s ) Γ ( 1 − s ) U − s ( − a ) \displaystyle M\left(\frac{1}{1+2a+{x}^{2}}\right) = \Gamma(s)\Gamma(1-s){U}_{-s}(-a)

Using Reflection Formula and remembering the definition of Chebyshev Polynomials,

= π s i n ( π s ) s i n ( ( 1 − s ) c o s − 1 ( − a ) ) s i n ( c o s − 1 ( − a ) ) \displaystyle = \frac{\pi}{sin(\pi s)} \frac{sin((1-s){cos}^{-1}(-a))}{sin({cos}^{-1}(-a))}

= π s i n ( π s ) s i n ( ( 1 − s ) c o s − 1 ( a ) ) 1 − a 2 \displaystyle = \frac{\pi}{sin(\pi s)} \frac{sin((1-s){cos}^{-1}(a))}{\sqrt{1-{a}^{2}}}

Check out this if this interested you!

Kartik Sharma - 5 years, 10 months ago

I've tried to approach it by English but I got stuck. Can you help?

I first start with a substitution of y = x 1 / 3 y = x^{1/3} then y = 2 z y = \sqrt2z . This simplifies the integral to

3 2 8 ∫ 0 ∞ d z ( z 3 + 1 4 ) 2 + ( 1 4 ) 2 \frac{3\sqrt2}8 \int_0^\infty \frac{dz}{\left(z^3+\frac14\right)^2 + \left(\frac14\right)^2}

Do you think that it's possible to proceed from here?

Pi Han Goh - 5 years, 10 months ago

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I believe yes, it is possible. Put on some substitution and then convert into sum so that Master theorem can be used. Actually Master theorem is nothing, just an extension to Gamma function applications. You should give it a read.

Kartik Sharma - 5 years, 10 months ago

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