Advanced Math/Complex integration
QUESTION: Hi Clyde, I am a research meteorologist brushing up on some complex analysis. I have a problem that I tried and worked out, but my calculator gives me a different answer and I can't figure out where I went wrong.
The problem is:
The integral of (x^2+x)/(x^5+1) dx from 0 to infinity.
I begin by solving x^5+1=0 and the three solutions I get in the upper half plane are approximately .81+.59i, -.31+.95i, and -1.
From here, I compute the residues at those three poles
I get the following residues
So then, I sum the residues and get -.143i so the integral is pi*i(-.143i)=.143*pi. My calculator gives me an answer of ~1.33. What am I missing?
ANSWER: First and foremost, how many solutions does x^5 + 1 = 0 have?
It has five, because it is a degree five polynomial. They are approximately:
x = -1
x ≈ 0.809017 + 0.587785 i
x ≈ 0.809017 - 0.587785 i
x ≈ -0.309017 - 0.951057 i
x ≈ -0.309017 + 0.951057 i
It is good to note that x=-1 is a removable singularity of this function. There is no residue there.
I don't know what kind of calculator you are using, but these residues are purely imaginary.
Res( (x^2+x)/(x^5+1) , 0.809017 + 0.587785 i ) ≈ -0.380423 i
Res( (x^2+x)/(x^5+1) , 0.809017 - 0.587785 i ) ≈ 0.235114 i
Your values are very poor -- off by over 1% in one case.
Now, I don't understand your choice of the upper half-plane for this integral. The integrand is not symmetric, and I think you are trying to divide your answer by 2 at the end to change the (real) integral from ( -∞ , ∞ ) to ( 0 , ∞ ). That won't work.
I would recommend using the first quadrant. This means your contours are, technically, going to have to be something like the segment [0,R], the segment [0,Ri] and the quarter circle from the real point R to the imaginary point Ri.
It doesn't seem like you care much to check for convergence, so I'll assume you can verify for yourself that the real integral [0,R] as R→∞ is the one you want (call this quantity capital I), while the integral along the imaginary contour [0,Ri] as R→∞ is going to be a multiple of I. If you do the substitution appropriately, you will get (1/4)(1+√5) I. (If you prefer, this is 0.809017 I.) The circular arc is easy to integrate, and bounded by 1/R^4 as R→∞, so that goes to zero.
So now the only residue in the first quadrant is -0.380423 i.
That should be equal to the sum of all three integrals, as R→∞, so:
I + 0.809017 I + 0 = 2 π i ( -0.380423 i )
1.809017 I = 2.39027
This gives you your integral, by dividing out:
I = 2.39027/1.809017 = 1.32131.
Note that all of these calculations can be done using exact values too. You'd obtain:
I = (2/5) √( 2 - 2/√5 ) π
---------- FOLLOW-UP ----------
QUESTION: I have one more question. What was the integral that gave you .809017? I thought I knew, but looking at it again, I'm confused.
It's not 0.80917, the integral is 0.80917 times the original integral. You can think of it as integrating the real variable from 0 to infinity, but the integral is ( (ix)^2 + (ix) ) / ( (ix)^5 + 1 ). You then relate that back to the original integral we want, which I called capital "I" above.
The choice of contour is not specific -- you can choose other contours, like the negative real axis or some other ray in some direction, so long as it does not intersect one of your non-removable singularities. You'd just have to account for any singularities inside that contour, which will depend on how you choose the contour in the first place.