Question #19134

integration from 0 to 1
of (m^(m-1)((1-x)^n-1)/ ((a+x)^(m+n)

Expert's answer

Conditions

integration from 0 to 1

of (m(m1)((1x)n1)/((a+x)(m+n))(m^{\wedge}(m - 1)((1 - x)^{\wedge}n - 1) / ((a + x)^{\wedge}(m + n))

Solution

mm1((1x)n1)(a+x)m+ndx=1m+n1mm1(1xa+1)n(a+x)mn+1((1xa+1)n(1x)2nF1(mn+1,n;mn+2;a+xa+1)+constant\begin{array}{l} \int \frac {m ^ {m - 1} ((1 - x) ^ {n} - 1)}{(a + x) ^ {m + n}} d x = \\ \frac {1}{m + n - 1} m ^ {m - 1} \left(\frac {1 - x}{a + 1}\right) ^ {- n} (a + x) ^ {- m - n + 1} \left(\left(\frac {1 - x}{a + 1}\right) ^ {n} - \right. \\ (1 - x) ^ {n} _ {2} F _ {1} \left(- m - n + 1, - n; - m - n + 2; \frac {a + x}{a + 1}\right) + \text{constant} \\ \end{array}mm1(a+x)mn+1m+n11m+n1mm1(a+1)n(a+x)2mn+1F1(mn+1,n;mn+2;a+xa+1)exp(2iπn[arg(a+1)2πarg(1x)2π+12])+constant\begin{array}{l} \frac {m ^ {m - 1} (a + x) ^ {- m - n + 1}}{m + n - 1} - \frac {1}{m + n - 1} m ^ {m - 1} (a + 1) ^ {n} \\ (a + x) ^ {- m - n + 1} _ {2} F _ {1} \left(- m - n + 1, - n; - m - n + 2; \frac {a + x}{a + 1}\right) \\ \exp \left(- 2 i \pi n \left[ \frac {\arg (a + 1)}{2 \pi} - \frac {\arg (1 - x)}{2 \pi} + \frac {1}{2} \right]\right) + \text{constant} \\ \end{array}

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