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This integral resisted all of my efforts to integrate it by parts. No matter how I improved the integrand, none of my ploys solved it: they either led me in circles (bringing me back to the integral I wanted to solve) or they led me into dead ends (taking me into situations in which I had to know the solution of the integral in order to solve it). I had to try something new, so I rewrote the cosine in terms of its imaginary exponential representation (from Euler's formula) and proceeded from there.
Once I made the substitution for the cosine, I multiplied both the numerator and the denominator of the integrand by eix and rearranged the denominator to obtain
Next I added to the denominator as if I were completing the square in preparation to solve it as a quadratic equation:
I factored the denominator:
I multiplied and divided the integral by twice the term that has the same sign in both of the factors of the denominator:
I split the numerator into equal parts and split an implied zero into the sum of the positive and negative imaginary terms in the denominator and then canceled out the common terms in the numerator and denominator, thereby obtaining two integrals:
I multiplied and divided each integral by the coefficient of eix in its denominator and by to convert its numerator into a proper differential of its denominator, simplified the leading coefficient, and carried out the integrations:
I exploited the fact that lnA-lnB = ln(A/B) and then multiplied both numerator and denominator of the combined logarithms' argument by :
I multiplied the numerator and denominator of the logarithm's argument by e-ix/2, converted the exponentials into their equivalent trigonometric functions via Euler's formula, and then divided both the numerator and the denominator by cos(x/2):
I wanted to simplify the argument of the logarithm, so I gathered together the real and imaginary terms in the numerator and the denominator and the factored out of the real terms and out of the imaginary terms:
Now I must figure out how I went wrong. I know that Equation 9 does not give me the correct solution of the integral because the integral should give me a real-valued solution and, for arbitrary values of a and b, Equation 9 does not yield a real-valued solution. However, if I multiply the first factor inside the bracket by its complex conjugate and multiply and divide the other factor by , I get
which corresponds to the solution in the Handbook of Chemistry and Physics. I appear to have cheated in making that move, but in fact I have merely added to Equation 6 a constant of integration,
and that's a legitimate move, justified by the necessity of making the solution a real-valued function of x and by the fact that it consists only of constants.
Finally, we have another way of representing the solution in Equation 10. We merely reverse the constants a and b and get
So we have
In that last move I absorbed the term coming from the iπ factor into the constant of integration, which I don't show explicitly (in accordance with standard practice in tables of integrals).
Average Radius of an Elliptical Orbit
If I want to make an accurate model of an orbit by mounting wire spokes on a hub representing the orbit's prime focus and if I want to put the same angle between all adjacent spokes, then in order to calculate the total length of wire I will need for my model I must calculate the average radius of the orbit and then multiply it by the number of spokes in the model. In calculating the average radius we can exploit the orbit's symmetry and carry out the integration over only half of the orbit, from peritelion to apotelion. We calculate the average radius of the orbit, then, by integrating the orbit's radius with respect to angular displacement, from θ=0 to θ=π and then divide the result by pi radians; that is, we have
We describe the radius of an orbit as a function of the traversed angle as
in which e represents the orbit's eccentricity and r0 represents the radial distance between the orbit's prime focus and its peritelion. Thus we have
From that equation we can readily calculate the amount of wire our model requires.
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