Can someone explain this mathematical "trick"? (This comes from a long derivation from pricing European options in Finance).
$$I = {1 \over {\sqrt {2\pi } }}\int\limits_{ - \infty }^{ - {x \over {\sqrt {2\tau } }}} {{e^{ - {1 \over 2}{{\left( {x' - {{(k - 1)\sqrt {2\tau } } \over 2}} \right)}^2}}}{e^{{1 \over 2}(k - 1)x + {1 \over 4}{{(k - 1)}^2}\tau }}} dx'$$
$$ = {{{e^{{1 \over 2}(k - 1)x + {1 \over 4}{{(k - 1)}^2}\tau }}} \over {\sqrt {2\pi } }}\int\limits_{ - \infty }^{ - {x \over {\sqrt {2\tau } }} - {1 \over 2}(k - 1)\sqrt {2\tau } } {{e^{-{1 \over 2}{s^2}}}ds} $$
$$={e^{{1 \over 2}(k - 1)x + {1 \over 4}{{(k - 1)}^2}\tau }}N( - d2)$$
I don't get how he can change the integral boundary? Also, how is that $N(-d2)$; when looking at the formula below, I don't see it, but help me please.
note: $$N(d) = {1 \over {\sqrt {2\pi } }}\int\limits_{ - \infty }^d {{e^{ - {1 \over 2}{s^2}}}ds} $$
$$d2 = {x \over {\sqrt {2\tau } }} + {1 \over 2}(k - 1)\sqrt {2\tau } $$
For the first part it looks like they made the substitution
$$ s = x' - \frac{1}{2}(k-1)\sqrt{2\tau} $$
Then, to get the new upper limit of integration you let $x' = -\frac{x}{\sqrt{2\tau}}$, which was the old limit. The lower limit does not change because $s \to -\infty$ as $x \to -\infty$. Note that the integrand after the substitution should actually be $e^{-\frac{1}{2}s^2}$.
As for the part with $N(-d2)$ , just replace $d$ with $-d2$ in the formula for $N(d)$ and you should see how the result matches your expression for $I$.