I Was Very Amazed At The Solution 6

For $a,b>0$ ,

$\displaystyle \begin{aligned} \int_{-\infty}^{\infty} e^{bx-ax^2}\; dx &= e^{b^2/4a}\int_{-\infty}^{\infty} e^{-a\left(x-\dfrac{b}{2a}\right)^2} \\ &= e^{b^2/4a}\sqrt{\dfrac{\pi}{a}}\end{aligned}$

where the last line follows from the result of a generalized Gaussian integral. Now ,

$\displaystyle \begin{aligned} \int_{-\infty}^{\infty} e^{-ax^2}\cos(bx)\; dx&=\frac12\left( \int_{-\infty}^{\infty} e^{-ax^2+ibx}\;dx + \int_{-\infty}^{\infty} e^{-ax^2-ibx}\; dx\right) \\ &= \int_{-\infty}^{\infty} e^{-ax^2+ibx}\; dx \\ &= \dfrac{e^{(ib)^2/4a}\sqrt{\pi}}{\sqrt{a}}\end{aligned}$

Thus we have ,

$\displaystyle \int_{0}^{\infty} e^{-ax^2}\cos(bx)\; dx=\dfrac{1}{2e^{b^2/4a}}\sqrt{\dfrac{\pi}{a}}$

Putting in $a=1,b=10$ we have the answer as $\displaystyle I=\dfrac{\sqrt{\pi}}{2e^{25}}$ making the answer $\boxed{25}$

Let $\displaystyle F(\alpha) = \int_0^{\infty} e^{-x^2} \cos(\alpha x) dx$

By Differentiating under the Integral Sign with respect to $\alpha$ ,

$\displaystyle \implies \dfrac{dF}{d\alpha} = \int_0^{\infty} e^{-x^2} \cdot - x\sin(\alpha x) dx$

Integrating by parts, that is $\displaystyle \int f'(x)g(x) = f(x)g(x) - \int f(x)g'(x)$ with $f'(x) = -xe^{-x^2} \implies f(x) = \dfrac{e^{-x^2}}{2} \ , \ g(x) = \sin(\alpha x) \implies g'(x) = \alpha \cos(\alpha x)$

$\displaystyle \implies \dfrac{dF}{d\alpha} = \left[\dfrac{e^{-x^2}}{2} \cdot \sin(\alpha x) \right]_0^{\infty} - \dfrac{\alpha}{2} \cdot \int_0^{\infty} e^{-x^2} \cos(\alpha x) dx$

$\displaystyle \dfrac{e^{-x^2}}{2} \cdot \sin(\alpha x)$ will approach 0 at both limits, hence we are left with:

$\displaystyle \dfrac{dF}{d\alpha} = - \dfrac{\alpha}{2} F \\ \implies \dfrac{dF}{F} = - \dfrac{\alpha}{2} d\alpha \\ \implies \ln(F) = - \dfrac{\alpha ^2}{4} + C \implies F(\alpha) = ge^{\frac{-\alpha ^2}{4}} \ , \ g = e^C$

Setting $\alpha = 0$

$\displaystyle \implies F(0) = \int_0^{\infty} e^{-x^2} dx = g = \dfrac{\sqrt{\pi}}{2} \implies F(\alpha) = \dfrac{\sqrt{\pi}}{2}e^{\frac{-\alpha ^2}{4}}$

Setting $\alpha = 10$

$\implies F(10) = I = \int_0^{\infty} e^{-x^2} \cos(10 x) dx = \dfrac{\sqrt{\pi}}{2}e^{-25} \\ \therefore \left|\ln\left(\dfrac{2I}{\sqrt{π}}\right)\right| = \boxed{25}$