Integral with exponential Brownian motion

Solution 1:

For the general case, I have not been able to establish the antiderivative required for computing $I$ which I rewrite $$I=\frac 1 {\sqrt{2 \pi}}\exp\Bigg[c \left(a+\frac{c T}{2}\right)\Bigg]\int_0^T t^{-\frac 32}\exp\Big[-\frac{a^2}{2 t}-\frac{c^2 t}{2}\Big]\,dt$$ For $J$ $$J=\frac 12 e^{a c}\exp\Big[\frac{c^2 T^2}{2}\Big]\Bigg[e^{-a c} \text{erfc}\left(\frac{a-c T}{\sqrt{2} \sqrt{T}}\right)+e^{a c} \text{erfc}\left(\frac{a+c T}{\sqrt{2} \sqrt{T}}\right) \Bigg]$$

Suppose $c=0$; for such a case, we have $$I=\frac 1 a \text{erfc}\left(\frac{a}{\sqrt{2T} }\right) \qquad \text{while} \qquad J=\text{erfc}\left(\frac{a}{\sqrt{2T} }\right) $$ Computing for various general cases, what it seems is that $$I =\frac J a $$ I supect (please confirm or negate) that your calculations have been done for $a=1$.