# Integral of an exponential that has a polynomial

How would one evaluate $$\Phi = \int_{-\infty}^{+\infty} e^{-(ax+bx^2)} dx$$.

I was trying to change it into a product of an error function and a gamma function, but I needed an extra dx. Any other ideas?

## Answers and Replies

Let X = ? such as : ax+bx² = A X² +B

I don't really see how that would work.

I tried setting u=ax, v=bx^2. Then i would use
$$\Gamma(1) = \int_{0}^{\infty} e^{-u}du$$
$${\rm erf}(x) = \int_{-\infty}^{x} e^{-v^2}dv$$
I'm not exactly sure how to use that in
$$\int_{-\infty}^{+\infty} e^{-u} e^{-v^2} du$$
without another differential dv somewhere.

You are not on the right way !
ax+bx² = b(x+(a/2b))² -a²/4b
X= x+(a/2b)
exp(-(ax+bx²)) = exp(a²/4b)*exp(-b X²)
Then integrate exp(-b X²)*dX

How would one evaluate $$\Phi = \int_{-\infty}^{+\infty} e^{-(ax+bx^2)} dx$$.

I was trying to change it into a product of an error function and a gamma function, but I needed an extra dx. Any other ideas?

$$\displaystyle{ \int_{-\infty}^{+\infty} e^{-(ax+bx^2)} dx= \int_{-\infty}^{+\infty} e^{-b\left(x+\frac{a}{2b}\right)^2}e^{\frac{a^2}{4b}} dx=\frac{e^{\frac{a^2}{4b}}}{\sqrt{b}} \int_{-\infty}^{+\infty} e^{-x^2} dx=e^{\frac{a^2}{4b}}\sqrt{\frac{\pi}{b}}\,,\,\,with}$$

(1) first equality: completing the square

(2) second equality: substituting $\displaystyle{\sqrt{b}\left(x+\frac{a}{2b}\right) \to y}$

(3) third equality: using $\displaystyle{\int_{-\infty}^\infty e^{-x^2}dx=\sqrt{\pi}}$

DonAntonio

Ps. Of course, I assume $b>0$