??a(\(\frac{u}{u}\)0)^(-b)(\(\frac{x}{x}\)0)^(-c)xdxd?
=a(\(\frac{u}{u}\)0)^(-b)(\(\frac{1}{x}\)0)^(-c)??x^(-c)xdxd?
??x^(-c)xdxd?=??x^(1-c)dxd?
=(?/8){(x2)^(2-c)-(x1)^(2-c)}/(2-c)

ｊｕｉｎさんの解答を詳しく書き直すと、

\(\int _{-\frac{\pi }{16}}^{\frac{\pi }{16}}\int _{x1}^{x2}a(\frac{u}{u0})^{-b}\cdot (\frac{x}{x0})^{-c}x\cdot dxd\theta\)

\(=a(\frac{u}{u0})^{-b}\cdot (\frac{1}{x0})^{-c}\int _{-\frac{\pi }{16}}^{\frac{\pi }{16}}\int _{x1}^{x2}(x)^{-c}\cdot x\cdot dxd\theta\)

二重積分のところのみ計算すると、

\(\int _{-\frac{\pi }{16}}^{\frac{\pi }{16}}\int _{x1}^{x2}\)

\(=\frac{(x2)^{^{2-c}}-(x1)^{^{2-c}}}{2-c}\int _{-\frac{\pi }{16}}^{\frac{\pi }{16}}d\theta\)

積分だけ計算すると、

\(\int _{-\frac{\pi }{16}}^{\frac{\pi }{16}}d\theta =\left[\)

したがって、答は

\(a(\frac{u}{u0})^{-b}\cdot (\frac{1}{x0})^{-c}\cdot \frac{(x2)^{^{2-c}}-(x1)^{^{2-c}}}{2-c}\cdot \frac{\pi }{8}\)

被積分函数
a(v/\(v_{0}\))^(-b) (x/\(x_{0}\))^(-c) x
には積分変数の θ が全く出て来ないので
θ に関しては定数。
従って先ず θ で積分してしまって
(π/8)∫_(\(x_{1}\))^(\(x_{2}\)) a(v/\(v_{0}\))^(-b) (x/\(x_{0}\))^(-c) x dx
= (π/8)a(v/\(v_{0}\))^(-b)(\(x_{0}\)\()^{c}\) ∫_(\(x_{1}\))^(\(x_{2}\)) x^(1-c) dx.
ここで
1) 1 - c = -1 即ち c = 2 の場合。(積分が収束するためには \(x_{1}\)・\(x_{2}\) > 0 が
必要)
与式 = (π/8)a(v/\(v_{0}\))^(-b)(\(x_{0}\)\()^{c}\) [log x]_(\(x_{1}\))^(\(x_{2}\))
= (π/8)a(v/\(v_{0}\))^(-b)(\(x_{0}\)\()^{c}\) log(\(x_{2}\)/\(x_{1}\)).

2) 1 － c ≠ -1 の場合。
与式 = (π/8)a(v/\(v_{0}\))^(-b)(\(x_{0}\)\()^{c}\) [(1/(2 - c))x^(2 - c)]_(\(x_{1}\))^(\(x_{2}\))
= (π/8)a(v/\(v_{0}\))^(-b)(\(x_{0}\)\()^{c}\) (1/(2 - c))((\(x_{2}\))^(2 - c) - (\(x_{1}\))^(2 - c)).