1+xn1+x^n in Fractions?

Could you calculate

∫11+xndx\large \int \frac{1}{1+x^n} dx

for every positive integer nn?

#Calculus

Note by Pepper Mint
3 years, 6 months ago

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1 vote

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Comments

Ok......What if someone replaces the + sign with a - sign........Can we solve it and generalize it??

Aaghaz Mahajan - 3 years, 3 months ago

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That is a good point we can use its series expansion... WAIT THE SERIES EXPANSION!!! We can use it to maybe SOLVE THE INTEGRAL (in series form BUT WHO CARES)!!! 11+x=1−x+x2−x3+x4−...=∑n=0∞(−1)nxn\frac { 1 }{ 1+x } =1-x+{ x }^{ 2 }-{ x }^{ 3 }+{ x }^{ 4 }-...=\sum _{ n=0 }^{ \infty }{ { (-1) }^{ n }{ x }^{ n } } 11+x2=1−x2+x4−x6+x8−...=∑n=0∞(−1)nx2n\frac { 1 }{ 1+{ x }^{ 2 } } =1-{ x }^{ 2 }+{ x }^{ 4 }-{ x }^{ 6 }+{ x }^{ 8 }-...=\sum _{ n=0 }^{ \infty }{ { (-1) }^{ n }{ x }^{ 2n } } 11+x3=1−x3+x6−x9+x12−...=∑n=0∞(−1)nx3n\frac { 1 }{ 1+{ x }^{ 3 } } =1-{ x }^{ 3 }+{ x }^{ 6 }-{ x }^{ 9 }+{ x }^{ 12 }-...=\sum _{ n=0 }^{ \infty }{ { (-1) }^{ n }{ x }^{ 3n } } So: 11+xk=1−xk+x2k−x3k+x4k−...=∑n=0∞(−1)nxkn\frac { 1 }{ 1+{ x }^{ k } } =1-{ x }^{ k }+{ x }^{ 2k }-{ x }^{ 3k }+{ x }^{ 4k }-...=\sum _{ n=0 }^{ \infty }{ { (-1) }^{ n }{ x }^{ kn } } and...: ∫11+xkdx=∫(1−xk+x2k−x3k+x4k−...)dx=∫∑n=0∞(−1)nxkndx\int { \frac { 1 }{ 1+{ x }^{ k } } dx } =\int { (1-{ x }^{ k }+{ x }^{ 2k }-{ x }^{ 3k }+{ x }^{ 4k }-...)dx } =\int { \sum _{ n=0 }^{ \infty }{ { (-1) }^{ n }{ x }^{ kn } } dx }

Simplifying the integral of the right: (Not including the +C in the integral) ∫∑n=0∞(−1)nxkndx=∑n=0∞∫(−1)nxkndx=∑n=0∞(−1)nxkn+1kn+1=x−xk+1k+1+x2k+12k+1−x3k+13k+1+x4k+14k+1−...\int { \sum _{ n=0 }^{ \infty }{ { (-1) }^{ n }{ x }^{ kn } } dx } =\sum _{ n=0 }^{ \infty }{ \int { { (-1) }^{ n }{ x }^{ kn }dx } } =\sum _{ n=0 }^{ \infty }{ \frac { { (-1) }^{ n }{ x }^{ kn+1 } }{ kn+1 } } =x-\frac { { x }^{ k+1 } }{ k+1 } +\frac { { x }^{ 2k+1 } }{ 2k+1 } -\frac { { x }^{ 3k+1 } }{ 3k+1 } +\frac { { x }^{ 4k+1 } }{ 4k+1 } -... ... which is so far all the work that can be done, without using any special functions.

Thus: ∫11+xkdx=∑n=0∞(−1)nxkn+1kn+1\int { \frac { 1 }{ 1+{ x }^{ k } } dx } =\sum _{ n=0 }^{ \infty }{ { \frac { { (-1) }^{ n }{ x }^{ kn+1 } }{ kn+1 } } } Done

Oon Han - 3 years, 2 months ago

Just glancing at it and some solutions computed with WolframAlpha, it looks like you have to use partial fractions to decompose it and then integrate term-by-term, which makes me unsure about whether or not a closed-form solution exists...

Aram Lindroth - 3 years, 3 months ago

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Integrate term by term... that is correct. In fact, the terms you need to integrate are actually very surprising.. x^n and -x^k! The answer is that ∫11+xkdx=∑n=0∞(−1)nxkn+1kn+1\int { \frac { 1 }{ 1+{ x }^{ k } } dx } =\sum _{ n=0 }^{ \infty }{ { \frac { { (-1) }^{ n }{ x }^{ kn+1 } }{ kn+1 } } } !

Oon Han - 3 years, 2 months ago

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You can see the power rule in the summation!

Oon Han - 3 years, 2 months ago

Anyway, we know that: ∫11+xdx=ln⁡(1+x)+C\int { \frac { 1 }{ 1+{ x } } } dx=\ln { (1+x) } +C and ∫11+x2dx=arctan⁡x+C\int { \frac { 1 }{ 1+{ x }^{ 2 } } } dx=\arctan { x } +C but ∫11+x3dx\int { \frac { 1 }{ 1+{ x }^{ 3 } } } dx is a mess...

(Just in case you don't believe me): ∫11+x3dx=−ln⁡∣x2−x+1∣−2(ln⁡∣x+1∣+3arctan⁡(2x−13))6+C\int { \frac { 1 }{ 1+{ x }^{ 3 } } } dx=-\frac { \ln { |{ x }^{ 2 }-x+1| } -2(\ln { |x+1| } +\sqrt { 3 } \arctan { (\frac { 2x-1 }{ \sqrt { 3 } } ) } ) }{ 6 } +C

Oon Han - 3 years, 3 months ago

@Pepper Mint Well, we can solve this sort of definite integral ranging from 0 to infinity.........this is simply using Beta function....!!!

Aaghaz Mahajan - 3 years ago

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But it won't help solve for the indefinite integral. hmm

Oon Han - 2 years, 12 months ago

@Pepper Mint Have a look at this paper...

Aaghaz Mahajan - 2 years, 2 months ago
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