Differentiate an exponential integral

Question

Would you guide me differentiating this integral for $m$:

$$\frac{\text{d}}{\text{d}m} \int_{x=-\infty}^m\int_{y=n}^{+\infty}\exp\left(-\left[\left(\frac{x-a}{b}\right)^2 - \frac{(x-a)(y-a)}{a}+\left(\frac{y-a}{c}\right)^2\right]\right)\space\text{d}y\text{d}x$$

noting that $a , b >0$

Answer 1

Ian say that $\frac{d}{dx} \int_a^x f(u) du = f(x)$ whatever a fixed.

so $\frac{d}{dm} \int_{x=-\infty}^m\int_{y=n}^{+\infty}\exp\left(-\left[\left(\frac{x-a}{b}\right)^2 - \frac{(x-a)(y-a)}{a}+\left(\frac{y-a}{c}\right)^2\right]\right)\space dy dx=\int_n^{\infty}\exp\left(-\left[\left(\frac{m-a}{b}\right)^2 - \frac{(m-a)(y-a)}{a}+\left(\frac{y-a}{c}\right)^2\right]\right)\space dy$.

Answer 2

$$I_1=\frac{\text{d}}{\text{d}m} \int_{x=-\infty}^m\int_{y=n}^{+\infty}\exp\left(-\left[\left(\frac{x-a}{b}\right)^2 - \frac{(x-a)(y-a)}{a}+\left(\frac{y-a}{c}\right)^2\right]\right)\space\text{d}y\text{d}x=$$ $$I_1=\frac{e^{-\frac{\left(4a^2-b^2c^2\right)(a-m)^2}{4a^2b^2}}\cdot\sqrt{\pi}\left(1+c\sqrt{\frac{1}{c^2}}\cdot\text{Erf}\left(\frac{a}{c}-\frac{c}{2}+\frac{cm}{2a}-\frac{n}{c}\right)\right)}{2\sqrt{\frac{1}{c^2}}}=$$ $$I_1=\frac{c\sqrt{\pi}e^{-\frac{\left(4a^2-b^2c^2\right)(a-m)^2}{4a^2b^2}}\left(1+\text{Erf}\left(\frac{2a^2-ac^2-2an+c^2m}{2ac}\right)\right)}{2}$$

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$$I_1=\frac{\text{d}}{\text{d}n} \int_{x=-\infty}^m\int_{y=n}^{+\infty}\exp\left(-\left[\left(\frac{x-a}{b}\right)^2 - \frac{(x-a)(y-a)}{a}+\left(\frac{y-a}{c}\right)^2\right]\right)\space\text{d}y\text{d}x=$$