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The Answer is: $\dfrac{e^{\frac{89}{8}}}{4}\sqrt{\dfrac{\pi}{2}}\left[\text{erfc}\left(\dfrac{3}{2\sqrt{2}}\right)+e^{-\frac{\pi^{2}}{2}}\text{Re}\left( i \text{erfc}\left(\dfrac{3-i2\pi}{2\sqrt{2}}\right)\right)\right]$ . Alright, Let's Begin!
First,we'll compute $I = \int_{0}^{\infty}e^{-(ax^{2}+bx+c)}\text{d}x$ Here, $a >0$ .
Setting $ax^{2}+bx=t$ . We get
$\text{d}t=(2ax+b)\text{d}x$
Also, $2ax+b = \sqrt{b^{2}+4at}$ . This came from the Quadratic Formula and the fact that as the limits are positive the positive square root is to be taken.
Now the changed integral is: $I=e^{-c}\int_{0}^{\infty}\dfrac{e^{-t}}{\sqrt{b^{2}+4at}}\text{d}t$ Again Letting $b^{2}+4at = z$
We get $\text{d}z=4a\text{d}t$ . Then the integral becomes: $I= \dfrac{e^{\frac{b^{2}-4ac}{4a}}}{4a}\int_{b^{2}}^{\infty} \dfrac{1}{\sqrt{z}}e^{-\frac{z}{4a}}\text{d}z$ Finally, we set $\frac{z}{4a} =p^{2}$ to obtain: $I=\dfrac{\sqrt{\pi}}{2\sqrt{a}}e^{\frac{b^{2}-4ac}{4a}}\text{erfc}\left(\dfrac{b}{2\sqrt{a}}\right)$ where the definition: $\text{erfc}(z) =\dfrac{2}{\sqrt{\pi}}\int_{z}^{\infty}e^{-x^{2}}\text{d}x$ was used.
Now coming to our main Integral, we Observe that: $\cos^{2}(\pi x) =\dfrac{1+\cos(2\pi x)}{2}$ Hence, we split the integral into two parts and evaluate them separately.
The First part comes directly from the above formula, and for the second, i.e. $J = 0.5\int_{0}^{\infty}\cos(2\pi x)e^{-(2x^{2}+3x-10)}\text{d}x$ We put $\cos(2\pi x)$ as $\text{Re}(e^{i2\pi x})$
This evaluation is left as an exercise for the reader, and henceforth, we get the solution by combining the integrals and after taking out all the possible common terms we are left with: $\dfrac{e^{\frac{89}{8}}}{4}\sqrt{\dfrac{\pi}{2}}\left[\text{erfc}\left(\dfrac{3}{2\sqrt{2}}\right)+e^{-\frac{\pi^{2}}{2}}\text{Re}\left( i \text{erfc}\left(\dfrac{3-i2\pi}{2\sqrt{2}}\right)\right)\right] \approx 3666.16$ . Also verified numerically, giving $A+B+C+D = 98$