Switching order of integration I faced this problem
$$\int_{\eta_0}^{\eta} (Bi(s) \int_\infty^s H(t) dt - Ai(s) \int_{\eta_0}^s H(t) dt )ds$$
How can I switch the order of integration? I am particularly interested on the first integral, since the interval (from infinity to s) is not clear to me. How would you do it?
I tried this, but I am not sure about the signs:
$$\int_{\eta_0}^{\eta} \int_{\eta_0}^t Bi(s)  H(t) ds dt + \int_{\eta}^{\infty} \int_{\eta_0}^\eta Bi(s)  H(t) ds dt -\int_{\eta_0}^{\eta} \int_{t}^\eta Ai(s) H(t)ds dt$$
 A: 
One possibility is to first split up the integration interval as $\int_{\infty}^{s}=\int_{\infty}^{\eta_{0}}+\int_{\eta_{0}}^{s}$. We find,
$$\begin{align}
I{\left(\eta,\eta_{0}\right)}
&=\int_{\eta_{0}}^{\eta}\mathrm{d}s\,\left[\operatorname{Bi}{\left(s\right)}\int_{\infty}^{s}\mathrm{d}t\,H{\left(t\right)}-\operatorname{Ai}{\left(s\right)}\int_{\eta_{0}}^{s}\mathrm{d}t\,H{\left(t\right)}\right]\\
&=\int_{\eta_{0}}^{\eta}\mathrm{d}s\,\left[\int_{\infty}^{s}\mathrm{d}t\,\operatorname{Bi}{\left(s\right)}\,H{\left(t\right)}-\int_{\eta_{0}}^{s}\mathrm{d}t\,\operatorname{Ai}{\left(s\right)}\,H{\left(t\right)}\right]\\
&=\int_{\eta_{0}}^{\eta}\mathrm{d}s\,\left[\int_{\infty}^{\eta_{0}}\mathrm{d}t\,\operatorname{Bi}{\left(s\right)}\,H{\left(t\right)}+\int_{\eta_{0}}^{s}\mathrm{d}t\,\operatorname{Bi}{\left(s\right)}\,H{\left(t\right)}-\int_{\eta_{0}}^{s}\mathrm{d}t\,\operatorname{Ai}{\left(s\right)}\,H{\left(t\right)}\right]\\
&=\int_{\eta_{0}}^{\eta}\mathrm{d}s\int_{\infty}^{\eta_{0}}\mathrm{d}t\,\operatorname{Bi}{\left(s\right)}\,H{\left(t\right)}+\int_{\eta_{0}}^{\eta}\mathrm{d}s\,\left[\int_{\eta_{0}}^{s}\mathrm{d}t\,\operatorname{Bi}{\left(s\right)}\,H{\left(t\right)}-\int_{\eta_{0}}^{s}\mathrm{d}t\,\operatorname{Ai}{\left(s\right)}\,H{\left(t\right)}\right]\\
&=-\int_{\eta_{0}}^{\eta}\mathrm{d}s\int_{\eta_{0}}^{\infty}\mathrm{d}t\,\operatorname{Bi}{\left(s\right)}\,H{\left(t\right)}+\int_{\eta_{0}}^{\eta}\mathrm{d}s\int_{\eta_{0}}^{s}\mathrm{d}t\,H{\left(t\right)}\left[\operatorname{Bi}{\left(s\right)}-\operatorname{Ai}{\left(s\right)}\right]\\
&=-\int_{\eta_{0}}^{\infty}\mathrm{d}t\int_{\eta_{0}}^{\eta}\mathrm{d}s\,H{\left(t\right)}\,\operatorname{Bi}{\left(s\right)}+\int_{\eta_{0}}^{\eta}\mathrm{d}t\int_{t}^{\eta}\mathrm{d}s\,H{\left(t\right)}\left[\operatorname{Bi}{\left(s\right)}-\operatorname{Ai}{\left(s\right)}\right]\\
&=-\int_{\eta_{0}}^{\infty}\mathrm{d}t\,H{\left(t\right)}\int_{\eta_{0}}^{\eta}\mathrm{d}s\,\operatorname{Bi}{\left(s\right)}+\int_{\eta_{0}}^{\eta}\mathrm{d}t\,H{\left(t\right)}\int_{t}^{\eta}\mathrm{d}s\,\left[\operatorname{Bi}{\left(s\right)}-\operatorname{Ai}{\left(s\right)}\right].\\
\end{align}$$
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