Integral Integral 60

The fact that $p_a (x)$ is a quadratic polynomial implies that some relation with the typical Gaussian integral can be established. Let $p_a (x) = ax^2 + bx + c$ to be general. We know that $x^2 + \frac{b}{a} x + \frac{c}{a} = (x + \frac{b}{2a})^2 + \frac{c}{a} - \frac{b^2}{4a^2}$ by the typical completion of the square, and if we multiply by $a$ we get that $p_a = ( \sqrt{a}x + \frac{b}{2\sqrt{a}})^2 + c - \frac{b^2}{4a}$ . With that established we can compute:

$\int_{- \infty}^{\infty} \frac{1}{e^{p_a (x)}} dx = \int_{-\infty}^{\infty} e^{-p_a(x)} dx = e^{\frac{b^2}{4a} - c} \int_{-\infty}^{\infty} e^{-(\sqrt{a} x + \frac{b}{2\sqrt{a}} )^2} dx$ . To compute this, let $u = \sqrt{a} x + \frac{b}{2\sqrt{a}}$ so $du = \sqrt{a} dx$ and we get $\frac{e^{\frac{b^2}{4a} - c}}{\sqrt{a}} \int_{-\infty}^{\infty} e^{-u^2}du = \frac{e^{\frac{b^2}{4a} - c}}{\sqrt{a}} \sqrt{\pi}$ .

Before performing the next integration, a simplification is key. The exponent of $e$ is $\frac{b^2}{4a} - c = \frac{b^2 - 4ac}{4a} = \frac{a}{4a} = \frac{1}{4}$ so what we have is actually just $\frac{e^{\frac{1}{4}}}{\sqrt{a}} \sqrt{\pi}$ . This means we only have to compute $\int_0^1 \frac{1}{\sqrt{a}} da = \int_0^1 a^{-\frac{1}{2}} da = 2$ to find that the answer to the problem is: $\sqrt{\pi} e^{\frac{1}{4}} 2$