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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

[an error occurred while processing this directive]---------- 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.

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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