Doubt on integration

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

#Calculus

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

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

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

$\begin{aligned} \displaystyle \int_0^1 \frac{1-x^3}{1-x^5} \, dx &=& \displaystyle \int_0^1 \displaystyle \sum_{r=0}^\infty (1-x^3) x^{5r} \, dx \\ \displaystyle &=& \sum_{r=0}^\infty \int_0^1 (x^{5r} - x^{3+5r} ) \, dx \\ \displaystyle &=& \sum_{r=0}^\infty \left( \frac1{5r+1} - \frac1{5r+4}\right ) \\ \displaystyle &=& \sum_{r=0}^\infty \frac3{25r^2+25r+4} \qquad (\star) \\ \end{aligned}$

We consider the logarithmic differentiation of the Weiestrass Product:

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

to get

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

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

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

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

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

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

Pi Han Goh - 5 years, 10 months ago

Is it correct to switch the integration and summation?

Pranshu Gaba - 5 years, 10 months ago

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

Pi Han Goh - 5 years, 10 months ago

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

Aditya Kumar - 5 years, 10 months ago

Great solution. Thanks.

Shivam Jadhav - 5 years, 10 months ago

What makes you think that it has an exact form?

Pi Han Goh - 5 years, 10 months ago

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

Shivam Jadhav - 5 years, 10 months ago

@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 - 5 years, 10 months ago

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

Pi Han Goh - 5 years, 10 months ago

yes it actually complicates it.

Aditya Kumar - 5 years, 10 months ago

Can you please solve it for me?

Shivam Jadhav - 5 years, 10 months ago

@Calvin Lin @Tanishq Varshney @Sandeep Bhardwaj

Shivam Jadhav - 5 years, 10 months ago

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

Calvin Lin Staff - 5 years, 10 months ago

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

Pi Han Goh - 5 years, 10 months ago

Hint: Factorize the denominator into two irreducible quadratic equations.

Pranshu Gaba - 5 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$