College Math Teaching

March 12, 2018

And I embarrass myself….integrate right over a couple of poles…

Filed under: advanced mathematics, analysis, calculus, complex variables, integrals — Tags: — collegemathteaching @ 9:43 pm

I didn’t have the best day Thursday; I was very sick (felt as if I had been in a boxing match..chills, aches, etc.) but was good to go on Friday (no cough, etc.)

So I walk into my complex variables class seriously under prepared for the lesson but decide to tackle the integral

\int^{\pi}_0 \frac{1}{1+sin^2(t)} dt

Of course, you know the easy way to do this, right?

\int^{\pi}_0 \frac{1}{1+sin^2(t)} dt =\frac{1}{2}  \int^{2\pi}_0 \frac{1}{1+sin^2(t)} dt and evaluate the latter integral as follows:

sin(t) = \frac{1}{2i}(z-\frac{1}{z}), dt = \frac{dz}{iz} (this follows from restricting z to the unit circle |z| =1 and setting z = e^{it} \rightarrow dz = ie^{it}dt and then obtaining a rational function of z which has isolated poles inside (and off of) the unit circle and then using the residue theorem to evaluate.

So 1+sin^2(t) \rightarrow 1+\frac{-1}{4}(z^2 -2 + \frac{1}{z^2}) = \frac{1}{4}(-z^2 + 6 -\frac{1}{z^2}) And then the integral is transformed to:

\frac{1}{2}\frac{1}{i}(-4)\int_{|z|=1}\frac{dz}{z^3 -6z +\frac{1}{z}} =2i \int_{|z|=1}\frac{zdz}{z^4 -6z^2 +1}

Now the denominator factors: (z^2 -3)^2 -8  which means z^2 = 3 - \sqrt{8}, z^2 = 3+ \sqrt{8} but only the roots z = \pm \sqrt{3 - \sqrt{8}} lie inside the unit circle.
Let w =  \sqrt{3 - \sqrt{8}}

Write: \frac{z}{z^4 -6z^2 +1} = \frac{\frac{z}{((z^2 -(3 + \sqrt{8})}}{(z-w)(z+w)}

Now calculate: \frac{\frac{w}{((w^2 -(3 + \sqrt{8})}}{(2w)} = \frac{1}{2} \frac{-1}{2 \sqrt{8}} and \frac{\frac{-w}{((w^2 -(3 + \sqrt{8})}}{(-2w)} = \frac{1}{2} \frac{-1}{2 \sqrt{8}}

Adding we get \frac{-1}{2 \sqrt{8}} so by Cauchy’s theorem 2i \int_{|z|=1}\frac{zdz}{z^4 -6z^2 +1} = 2i 2 \pi i \frac{-1}{2 \sqrt{8}} = \frac{2 \pi}{\sqrt{8}}=\frac{\pi}{\sqrt{2}}

Ok…that is fine as far as it goes and correct. But what stumped me: suppose I did not evaluate \int^{2\pi}_0 \frac{1}{1+sin^2(t)} dt and divide by two but instead just went with:

$latex \int^{\pi}_0 \frac{1}{1+sin^2(t)} dt \rightarrow i \int_{\gamma}\frac{zdz}{z^4 -6z^2 +1} where \gamma is the upper half of |z| = 1 ? Well, \frac{z}{z^4 -6z^2 +1} has a primitive away from those poles so isn’t this just i \int^{-1}_{1}\frac{zdz}{z^4 -6z^2 +1} , right?

So why not just integrate along the x-axis to obtain i \int^{-1}_{1}\frac{xdx}{x^4 -6x^2 +1} = 0 because the integrand is an odd function?

This drove me crazy. Until I realized…the poles….were…on…the…real…axis. ….my goodness, how stupid could I possibly be???

To the student who might not have followed my point: let \gamma be the upper half of the circle |z|=1 taken in the standard direction and \int_{\gamma} \frac{1}{z} dz = i \pi if you do this property (hint: set z(t) = e^{it}, dz = ie^{it}, t \in [0, \pi] . Now attempt to integrate from 1 to -1 along the real axis. What goes wrong? What goes wrong is exactly what I missed in the above example.


April 9, 2015

I’d be a better mathematician if I weren’t so freaking dumb!!!!

Filed under: academia, editorial — Tags: — collegemathteaching @ 7:45 pm

I have some time to think about things, and I got hung up on a detail…and the resolution of said detail was realizing, that if f is a bijection (one to one and onto), then (f^{-1})^{-1} = f .


Back to work.

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