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We solve this integral by the same technique that we used to solve the version that has a minus sign in the denominator, though in this case we will use complex numbers. We begin by factoring the denominator;
Next, multiply and divide the right side of the equation by 2a and then undo the implicit addition in the numerator;
That gives us the sum of two integrals;
Multiplying and dividing both of those integrals by ib and integrating them directly gives us
Now we have two options for putting that result into its final form. When we alter them to reflect the fact that I used squared coefficients a2 and b2, instead of the first powers of the coefficients, the solutions in the Handbook of Chemistry and Physics (64th Edition, 1983-1984) come to us as
I: We take our first option by acknowledging that the difference between two logarithms equals the logarithm of the ratio of their arguments. Thus, Equation 4 becomes
which corresponds to Equation 6.
II: We take our second option by multiplying and dividing the arguments of the logarithms in Equation 4 by , the magnitude of the complex numbers that comprise those arguments, then splitting the logarithms of the products into sums of the logarithms of the factors to obtain
Next we note that the division of the right side of that equation by i corresponds to taking the logarithms of the minus i-th power of the logarithms' arguments. To exploit that fact we need to put the arguments of the logarithms into complex polar form; that is, we must have
In this case we have β=0 and α=tan-1(bx/a), so we have for the minus i-th power of Equations 9 and 10
Now Equation 8 becomes
which matches Equation 5.
Illumination of a Moving Sphere
by a Point Source
Someone has sent a probe into intragalactic space at a speed, relative to the stars it passes, of between 10 and 50 percent of the speed of light. At that speed the probe passes its target stars more or less in a straight line, but without encountering any major relativistic effects. The probe has the form of a sphere of radius R and, to save weight, its builders designed it to capture and store the energy in starlight, through the photocells that cover its surface, as it passes near a star.
For simplicity of calculation we make the probe's path coincide with the x-axis of our coordinate grid and thus describe the probe's velocity as v=dx/dt, which we can treat as a constant. The probe makes its closest approach to a star at x=0, a point that lies a distance Y from the center of the star. The star emits energy at the rate dE/dt, so the intensity of the light striking the probe at any given instant equals (dE/dt)/r2, in which r2=Y2+x2.
In a given instant of time, dt, the probe absorbs energy in the amount
But dt=dx/v, so the integral that sums that differential comes to us as
We evaluate that integral between the points x=-x0, where the star's light becomes just dense enough to generate power at a rate the probe can use, and x=+x0, where the star's light dims to a density that the probe cannot use. We have, then,
If we make the ratio x0/Y greater than 20, then the arctangent comes close enough to π/2 that we can write, to a good approximation,
If you plot the integrand of the integral in Equation 16 as a z(x) versus x graph, you will obtain a curve similar to the curve known as the Witch of Agnesi. The name is actually a mistranslation. It derives from a book on the calculus written by Maria Gaetana Agnesi in 1748. She called it a versed-sine curve. Unfortunately, the Italian word for versed sine resembles one of the Italian words for witch and the English mathematician, John Colson, who translated Agnesi's book in 1801 was not sufficiently proficient in Italian to make the correct translation (It's as if a foreigner translating an English work in astrophysics translated a reference to neutral hydrogen (un-ionized) as hydrogen that had formed a labor union).
The Witch itself comprises the set of
points defined in the following way: On an x-z grid draw a circle of radius A
with its center at the point (0,A) and draw the two tangents to the circle, z=0
and z=2A. Establish a straight line that can pivot about the point (0,0). Draw
the line M parallel to the x-axis through the point where the pivot line crosses
the circle and draw the line N parallel to the z-axis through the point where
the pivot line crosses the tangent z=2A. The point where M and N intersect each
other is an element of the set of points comprising the Witch. The curve thus
defined satisfies the equation
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