∫

We would have a straightforward solution of this integral if we could multiply the integrand by x. We would then have

(Eq'n 1)

But if we multiply the integrand by x, then we must also divide it by x and then integrate the result by parts, a procedure that, in this case, leads us into an infinite progression.

Serendipity intervenes by showing us that, in addition to Equation 1, we have

(Eq'n 2)

Those two equations complement each other, so we can add them together to get

(Eq'n 3)

But that has the form of

(Eq'n 4)

which allows us to write

(Eq'n 5)

Thus we solve the integral.

Example Application:

Gravitational Deflection of Starlight-

the Refractive Component

A wave train of proper length L_{0}
flies past the sun on a path that we approximate with a straight line. At any
point on its flight path a distance x from its perihelion point (which lies a
distance R from the sun's center) each element of length becomes

(Eq'n 6)

in which the proper length dx obeys the
equation L_{0}=∫dx and we have

(Eq'n 7)

The total length of the wave train then comes to us as

(Eq'n 8)

In the second step of that integration I
replaced the radical in the integrand with the first two terms of its equivalent
Taylor expansion, something I can legitimately do only when we have R very much
greater than 2MG/c^{2} = 2.956 kilometers. The decrement from L_{0}
in that equation represents the extra length that the gravitational contraction
of space draws into the distance between two fixed points and thus gives us a
delay in the transmission of radio signals (Shapiro's delay).

If two identical wave trains fly side by side and pass through perihelia that differ by dR in the radial direction, then the requirement that the waves continue to fly side by side necessitates that the waves turn through an angle β given by Sinβ=dL'/dR. Because we have approximated the course by a straight line, we also have the approximation β = Sinβ. We have then

(Eq'n 9)

when x goes from -∞ to +∞.

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