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Reference: Batchelor, Introduction to Fluid Dynamics after eqn (6.9.8)

Assuming[{Element[n, Integers], n > 0, Element[a, Reals], a > 0, 
Element[t, Reals], t > 0}, 
Integrate[Cos[n*t]/(Cos[t] - Cos[a]), {t, 0, Pi}, PrincipalValue -> True]]

A "standard" definite integral Pi* Sin[n a]/Sin[a]. However, I can't get Mathematica to show equality between its result and this. And it is not an obvious listing In Gradshetyn either. Batchelor did prove his result in a footnote from recursion after an indefinte integration for n = 0

Log[Abs[Sin[(a + t)/2]/Sin[(a - t)/2]]]/Sin[a]

which Mathematica does do. Please include successful code for the definite integrals. I was going to try Mathematica extending the theory to oscillating lift but a prerequisite is that it can do the steady case easily and not recursively.

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    $\begingroup$ I don't see any clear statement of what you are asking us to do to help you. Could you please edit the question to provide that? $\endgroup$ – m_goldberg Feb 10 at 16:24
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    $\begingroup$ For n=0 Mathematica finds $I_0=i\pi \csc {a}$ so the real part is 0. For n=1, Mathematica finds $I_1=\pi - i\pi \cot {a}$, so the real part is $\pi $, as well as in Batchelor 's book. Then use the recurrent formula. $\endgroup$ – Alex Trounev Feb 10 at 19:48
  • $\begingroup$ Please post your successful code as I am still have trouble with the definite integrals in Mathematica. $\endgroup$ – simon Feb 10 at 21:24
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1) $n=0$

Integrate[1/(Cos[t] - Cos[a]), {t, 0, Pi}, PrincipalValue -> True]
(*ConditionalExpression[
 I \[Pi] Csc[
   a], (Re[ArcCos[Cos[a]]] < 0 || Re[ArcCos[Cos[a]]] > \[Pi] || 
    ArcCos[Cos[a]] \[NotElement] Reals) && Sin[a/2] >= 0 && 
  Cos[a/2] <= 0]*)

2)$n=1$

Integrate[Cos[t]/(Cos[t] - Cos[a]), {t, 0, Pi}, 
 PrincipalValue -> True]
(*ConditionalExpression[\[Pi] - I \[Pi] Cot[a], 
 Cos[a/2] >= 
   0 && (Re[ArcCos[Cos[a]]] < 0 || Re[ArcCos[Cos[a]]] > \[Pi] || 
    ArcCos[Cos[a]] \[NotElement] Reals) && Sin[a/2] <= 0]*)

We are interested in the real part of the resulting expressions: $n=0, I_0=0$ and $n=1, I_1=\pi $ Thus, the real part of the resulting expressions is the same as in the book. Then use the recurrent formula $I_{n+1}+I_{n-1}=2I_n\cos{a}$

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  • $\begingroup$ It took awhile to see how to read these code answers and ignore the inserted backslashes and Mathematica's conditions. They are simpler than what v 7 gave me and allow me to see its Re part was also zero. Ignore. Ia >= [Pi] it had I0 =Csc[a] (I*Pi + Log[2] - Log[-2 Cos[a/2]] + Log[Cos[a/2]]) $\endgroup$ – simon Feb 11 at 1:18

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