Doubt on integration

How to solve this question 01x2+x+1x4+x3+x2+x+1dx\int\limits_0^1\frac{x^{2}+x+1}{x^{4}+x^{3}+x^{2}+x+1}dx. Please help me in solving this.

Note by Shivam Jadhav
4 years ago

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Multiply top and bottom by (x1)-(x-1), we have 011x31x5dx \displaystyle \int_0^1 \frac{1-x^3}{1-x^5} \, dx .

Notice that the integrand can be written as a sum of a convergent geometric series with common ratio x5x^5.

011x31x5dx=01r=0(1x3)x5rdx=r=001(x5rx3+5r)dx=r=0(15r+115r+4)=r=0325r2+25r+4()\begin{aligned} \displaystyle \int_0^1 \frac{1-x^3}{1-x^5} \, dx &=& \displaystyle \int_0^1 \displaystyle \sum_{r=0}^\infty (1-x^3) x^{5r} \, dx \\ \displaystyle &=& \sum_{r=0}^\infty \int_0^1 (x^{5r} - x^{3+5r} ) \, dx \\ \displaystyle &=& \sum_{r=0}^\infty \left( \frac1{5r+1} - \frac1{5r+4}\right ) \\ \displaystyle &=& \sum_{r=0}^\infty \frac3{25r^2+25r+4} \qquad (\star) \\ \end{aligned}

We consider the logarithmic differentiation of the Weiestrass Product:

cos(πx)=n=0(14x2(2n+1)2)\displaystyle \cos(\pi x) = \prod_{n=0}^\infty \left( 1 - \frac{4x^2}{(2n+1)^2} \right)

to get

n=01(2n+1)2(2x)2=π8xtan(πx)() \displaystyle \sum_{n=0}^\infty \frac1{(2n+1)^2 - (2x)^2} = \frac{\pi}{8x} \tan(\pi x ) \qquad (\star \star)

Notice that 325r2+25r+4=12251(2r+1)2(2310)2 \frac3{25r^2+25r+4} = \frac{12}{25} \cdot \frac1{(2r+1)^2 - \left (2\cdot \frac3{10} \right)^2} . Thus x=310x=\frac3{10} for ()(\star \star) .

Hence, the integral in question equals to 1225π83/10tan(π310)=π5tan(310π) \dfrac{12}{25} \cdot \dfrac{\pi}{8\cdot 3/10} \tan\left(\pi \cdot \dfrac3{10}\right) = \dfrac\pi5 \tan\left(\dfrac3{10}\pi\right) .

Now, we just need to evaluate tan(310π)\tan\left(\frac3{10}\pi\right). Apply the identity tan(x)=1cos(2x)1+cos(2x) \tan(x) = \sqrt{\frac{1-\cos(2x)}{1+\cos(2x)}} for x=310πx = \frac3{10}\pi .

This means we need to find what is the value of cos(35π) \cos\left(\frac35\pi \right) . Let yy denote this value. Because cos(15π)=cos(4×15π) \cos\left(\frac{1}5\pi\right) = -\cos\left( 4\times \frac15\pi \right) . Apply the double angle formula twice yields cos(15π)=1+54\cos\left( \frac15\pi\right) = \frac{1+\sqrt5}4 . Apply the triple angle formula to get y=154y = \frac{1-\sqrt5}4 .

Substitution yields the answer of π5+251250.8648  \displaystyle \pi \sqrt{\dfrac{5+2\sqrt5}{125}} \approx 0.8648 \ \square

Pi Han Goh - 4 years ago

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Is it correct to switch the integration and summation?

Pranshu Gaba - 4 years ago

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Yes, apply Fubini's (or Tonelli's) theorem.

Pi Han Goh - 4 years ago

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@Pi Han Goh Can u please lend me ur brains for giving jee. Lol i was just kidding. Amazing sol.

Aditya Kumar - 4 years ago

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Great solution. Thanks.

Shivam Jadhav - 4 years ago

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What makes you think that it has an exact form?

Pi Han Goh - 4 years ago

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This question was given to me in my classes and was also told that answer was in terms of pi.

Shivam Jadhav - 4 years ago

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@Shivam Jadhav divide the denominator by the numerator. Then u can split it up into partial fractions. Then u can proceed it using normal methods

Aditya Kumar - 4 years ago

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Wait what? How is that possible? Doesn't it complicate things?

Pi Han Goh - 4 years ago

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yes it actually complicates it.

Aditya Kumar - 4 years ago

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Can you please solve it for me?

Shivam Jadhav - 4 years ago

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Check out the Integration of Rational Functions that @Pranshu Gaba and @Vishnuram Leonardodavinci have been working on.

Calvin Lin Staff - 4 years ago

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Answer is π5+251250.8648 \pi \sqrt{\dfrac{5+2\sqrt5}{125}} \approx 0.8648 . I will post the solution shortly.

Pi Han Goh - 4 years ago

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Hint: Factorize the denominator into two irreducible quadratic equations.

Pranshu Gaba - 4 years ago

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