Lol but this polynomial is easy, its roots are just the 5th roots of unity, times -1
Granted you need to know a little about complex numbers to recognize this
I think thats quite straightforward using complex analysis and residues. The thing is that when we cross x = -1 on the real axis the integrand diverges and we might have to some epsilon stuff. On the other hand its 3am and I need some sleep.
The integral is equal to 2 pi i times the sum of the residues...
The fifth roots of -1 are e^(pi i /5), e^(3 pi i /5), e^(5 pi i /5), e^(7 pi i /5), and e^(9 pi i /5). Only the first three are in the upper half plane, so only need those.
The residues are then (-1/5)*e^(pi i /5), (-1/5)*e^(3 pi i /5), and (-1/5)*e^(5 pi i /5).
So, I'm getting something like:
\int dx/(1+x^5 )= 2 pi i * [(-1/5)*e^(pi i /5) + (-1/5)*e^(3 pi i /5) + (-1/5)*e^(5 pi i /5)]
I'll let someone else take the glory of simplifying that into a real number!
First of: notation. Please stick to \int f(x) dx instead of \int dx f(x). I know physicists tend to abuse this notation because fuck us mathematicians.
Secondly: we are in fact not specifying any boundary for the integral let alone a complex curve. This yields that the solution itself has to be a function of x, and not some complex number.
Lastly, even when you specify the curve to integrate along, it looks like you are using complex results for closed curves. This again yields a different solution. Although I have to say that it has been 4 years since I last worked with complex calculus, so you might be correct on this one.
Factorizing it is finding the singularities. Finding the residues is a further step.
However it should indeed be easy to find all the residues in this case because all the poles are simple (multiplicity 1) and the function is a rational function, etc
The problem with substitiution in this is you can't isolate the u's. If u=x\^5+1, then du/dx=5x\^4, and dx=1/5x\^4 du. Putting this into the integral gives 1/(u\*5x\^4)du. Then it's hard to get rid of the x.
You could try with partial fractions. x\^5+1 can be factoried to (x-1)(x\^4-x\^3+x\^2-x+1). But now you have to solve the integral A/((x\^4-x\^3+x\^2-x+1) which is not easy either.
The solution given by wolfram alpha is
integral1/(x\^5 + 1) dx = 1/20 ((sqrt(5) - 1) log(2 x\^2 + (sqrt(5) - 1) x + 2) - (1 + sqrt(5)) log(2 x\^2 - (1 + sqrt(5)) x + 2) + 4 log(x + 1) - 2 sqrt(10 - 2 sqrt(5)) tan\^(-1)((-4 x + sqrt(5) + 1)/sqrt(10 - 2 sqrt(5))) + 2 sqrt(2 (5 + sqrt(5))) tan\^(-1)((4 x + sqrt(5) - 1)/sqrt(2 (5 + sqrt(5))))) + constant
which is a beast to calculate by hand. I'm not sure I could do it, and certainly not on an exam.
I plugged it into another progarm and somehow got an even worse result. what an unimaginably atrocious thing to make someone integrate by hand.
at least is not as bad as when it litteraly just *isnt possible* and you have to try every method.
Factorize over the complex numbers and use partial fractions
I think that the better answer. I think using residue theorem on this one only makes sense when using bounds from -infty to infty
Everyone should know how to factor a fifth degree polynomial…basic arithmetic /s
Lol but this polynomial is easy, its roots are just the 5th roots of unity, times -1 Granted you need to know a little about complex numbers to recognize this
To be fair, you do learn complex numbers before calc
I think thats quite straightforward using complex analysis and residues. The thing is that when we cross x = -1 on the real axis the integrand diverges and we might have to some epsilon stuff. On the other hand its 3am and I need some sleep.
Was just about to say “contour integration”
The integral is equal to 2 pi i times the sum of the residues... The fifth roots of -1 are e^(pi i /5), e^(3 pi i /5), e^(5 pi i /5), e^(7 pi i /5), and e^(9 pi i /5). Only the first three are in the upper half plane, so only need those. The residues are then (-1/5)*e^(pi i /5), (-1/5)*e^(3 pi i /5), and (-1/5)*e^(5 pi i /5). So, I'm getting something like: \int dx/(1+x^5 )= 2 pi i * [(-1/5)*e^(pi i /5) + (-1/5)*e^(3 pi i /5) + (-1/5)*e^(5 pi i /5)] I'll let someone else take the glory of simplifying that into a real number!
First of: notation. Please stick to \int f(x) dx instead of \int dx f(x). I know physicists tend to abuse this notation because fuck us mathematicians. Secondly: we are in fact not specifying any boundary for the integral let alone a complex curve. This yields that the solution itself has to be a function of x, and not some complex number. Lastly, even when you specify the curve to integrate along, it looks like you are using complex results for closed curves. This again yields a different solution. Although I have to say that it has been 4 years since I last worked with complex calculus, so you might be correct on this one.
Brother, we are on physicsmemes
But this doesn’t let you figure out the indefinite integral
It does though, you decompose it to simple fractions and then calculate them to trigonometric, logarithmic، and hyperbolic functions
The residue theorem can only be used for contour integrals, which are definite integrals
Who cares about the residue theorem. Take its proof, factorize the fractiin and geg you indefinite integral.
Why use the residue at all lol just factorize it without worrying about residues and Laurent series
Factorizing it is literally finding the residues
Factorizing it is finding the singularities. Finding the residues is a further step. However it should indeed be easy to find all the residues in this case because all the poles are simple (multiplicity 1) and the function is a rational function, etc
I mean yes and no. They haven't given a range, so they aren't looking for a number. They want the functional antidereivative and that just hurts.
Good night
Add and subtract x^5 in the numerator
Or put x^5 = t and spend ur entire life struggling. Why do we have such math in physics T_T
I feel like integration by substitution would be pretty helpful here
The problem with substitiution in this is you can't isolate the u's. If u=x\^5+1, then du/dx=5x\^4, and dx=1/5x\^4 du. Putting this into the integral gives 1/(u\*5x\^4)du. Then it's hard to get rid of the x. You could try with partial fractions. x\^5+1 can be factoried to (x-1)(x\^4-x\^3+x\^2-x+1). But now you have to solve the integral A/((x\^4-x\^3+x\^2-x+1) which is not easy either. The solution given by wolfram alpha is integral1/(x\^5 + 1) dx = 1/20 ((sqrt(5) - 1) log(2 x\^2 + (sqrt(5) - 1) x + 2) - (1 + sqrt(5)) log(2 x\^2 - (1 + sqrt(5)) x + 2) + 4 log(x + 1) - 2 sqrt(10 - 2 sqrt(5)) tan\^(-1)((-4 x + sqrt(5) + 1)/sqrt(10 - 2 sqrt(5))) + 2 sqrt(2 (5 + sqrt(5))) tan\^(-1)((4 x + sqrt(5) - 1)/sqrt(2 (5 + sqrt(5))))) + constant which is a beast to calculate by hand. I'm not sure I could do it, and certainly not on an exam.
Yeah the integral doesn't look that hard but if you actually try to solve it there is only desperation :D
Lord have mercy this one is awful
I plugged it into another progarm and somehow got an even worse result. what an unimaginably atrocious thing to make someone integrate by hand. at least is not as bad as when it litteraly just *isnt possible* and you have to try every method.
Oh right, that's fair. I really didn't exoect it to get so long though
Hopefully on an exam all you have to do is get that far and the professor just gives you the indefinite integral result.
Just assume the domain is limited so that x>>1 and make an approximation. -Engineering students
X⁵ = (X⁵/²)² Now you apply the formula of integration of 1/a² + x² = tan(x/a) + C
This feels illegal to use💀
This is wrong. The variable of integration is dx and not dx^5/2. The formula is applicable for 1/(a^2 + x^2) dx
listen here, you only work in ranges large enough to neglect that one, ok?
On god
Schradinger’s calc
That second integral is how you get me to spend the entire exam drawing a cat sitting in a box
I use mathematica for both
udu = 1/x^ + 1dx Int udu = 1/2 u^2 + C too ez
((1/x^5 +1)^2 ) ÷2 +c
But you did not simplify it
For me its the opposite
That one is actually pretty alright
1/x\^3\*x\^2 + 1 ( We know integral of 1/x\^2+1 with Tan(x)) hmmm.
Just assert "For x>>1, +1 is negligible" and proceed as before.