Help with substitution

Question

I am having troubles with the following transformation. I have: \begin{align*} Y_t^k=\int_0^t(t-s)^{\alpha-1}h(s)ds+\int_0^t (t-s)^{\alpha-1}\int_0^s(s-u)^{-\alpha}\sigma_{n_k}(X_u^{n_k})dB_u^{n_k}ds \end{align*} and must show that this is equivalent to \begin{align*} Y_t^k=\int_0^t(t-s)^{\alpha-1}h(s)ds+ c_{\alpha}\int_0^{t}\sigma_{n_k}(X_u^{n_k})dB_u^{n_k}, \end{align*} where $c_\alpha =\int_0^1 (1-r)^{\alpha-1}r^{-\alpha}dr$. My ansatz looks as follows: Substitute $r=\frac{s}{t}$ to get \begin{align*} Y_t^k&=\int_0^t(t-s)^{\alpha-1}h(s)ds+\int_0^1 (t-rt)^{\alpha-1}\int_0^{rt}(rt-u)^{-\alpha}\sigma_{n_k}(X_u^{n_k})dB_u^{n_k}tdr\\ &=\int_0^t(t-s)^{\alpha-1}h(s)ds+\int_0^1 (t-rt)^{\alpha-1}\int_0^{rt}(rt-u)^{-\alpha}\sigma_{n_k}(X_u^{n_k})dB_u^{n_k}(\frac{1}{t})^{\alpha-1-\alpha}dr\\ &=\int_0^t(t-s)^{\alpha-1}h(s)ds+\int_0^1 (1-r)^{\alpha-1}\int_0^{rt}(r-\frac{u}{t})^{-\alpha}\sigma_{n_k}(X_u^{n_k})dB_u^{n_k}dr. \end{align*} But now I am clueless how to transform the incorrect term to the thing in $c_\alpha$ and the correct integral bound. I am very thankful for ideas!

Answer 1

After you switched the order of integration, you're left with the integral \begin{align} \int_u^t(t-s)^{\alpha-1}(s-u)^{-\alpha} ds. \end{align} In order to reduce the limits of the integral to 0 and 1, we just choose $r=(s-u)/(t-u)$. Now you substitute this in the integral and you are done. However, the general technique you are using here is called Factorization, which crucially depends on the identity \begin{align} \int_u^t(t-s)^{\alpha-1}(s-u)^{-\alpha} ds=\frac{\pi}{\sin(\alpha\pi)}. \end{align} Hence, as far as I am concerned you do not need to compute the transformation but just define $c_\alpha$ as $\frac{\pi}{\sin(\alpha\pi)}$.