I am unable to understand the following integral $$ \int Q(b+cx) dx = \frac{1}{c}\left[(b+cx)Q(b+cx)-\frac{1}{\sqrt{2\pi}}exp\{-\frac{(b+cx)^2}{2}\}\right] .......(1) $$
where Q(x) is defined as $$ Q(x) = \frac{1}{\sqrt{2\pi}}\int_x^\infty exp\{\frac{-y^2}{2}\}dy .....(2) $$
I know how to solve the following integral $$ \int Q(x)dx=xQ(x)-\frac{1}{\sqrt{2\pi}}exp\{\frac{-x^2}{2}\} ......(3) $$
Actually i need to solve $$ \int Q(b-cx)dx = ? ..... (4) $$
but for that I must understant the first equation.
kindly help me,
cordially
Because $d (b+cx)=c dx$
From (1) we obtain:
$$c\int Q(b+cx) dx=\int Q(b+cx) d(b+cx) =(b+cx)Q(b+cx)-\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{(b+cx)^2}{2}\right) .......(2)$$
Set $u=b+cx$ we have
$$\int Q(u) du =u Q(u)-\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{u^2}{2}\right) .......(3)$$
This is exactly what suggested by gammatester!
Similarly: $d (b-cx)=-c dx$
$$-c\int Q(b-cx) dx=\int Q(b-cx) d(b-cx) =(b-cx)Q(b-cx)-\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{(b-cx)^2}{2}\right) .......(4)$$