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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
Res(x)=(x^2+x)/5x^4=(x+1)/5x^3

I get the following residues
-.001-.378i,.001+.235i,0

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?

Thanks.

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.

Thanks.

Answer
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.

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Clyde Oliver

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I can answer all questions up to, and including, graduate level mathematics. I am more likely to prefer questions beyond the level of calculus. I can answer any questions, from basic elementary number theory like how to prove the first three digits of powers of 2 repeat (they do, with period 100, starting at 8), all the way to advanced mathematics like proving Egorov's theorem or finding phase transitions in random networks.

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