Calculating Integral for $(1+2^kw)^aD(y,z,w)$

[fderingoz]

we know that \int_{|y-z|}^{y+z}D(y,z,w)\frac{w^{2m+1}}{2^m\Gamma(m+1)}dw=1<br />
how can we calculate the integral
\int_{|y-z|}^{y+z}(1+2^kw)^aD(y,z,w)\frac{w^{2m+1}}{2^m\Gamma(m+1)}dw

 

[Whitishcube]

just a shot in the dark, but maybe integration by parts. if you can somehow make the first part of that integral disappear, you can finish it off.

 

[physeven]

integration by parts would be a method to use but the problem would be integrating the (1+2^kw)^a part but if you choose u(x) and v(x) appropriately you will solve the problem since you already know that the first integral with the same limits does equal 1.