In particular, I am asking about integrals shaped like this: $$\int_{-\infty}^\infty e^{-ax^2-bx}dx$$ where $a\in\mathbb C$ with ${\rm Re}(a)>0$, and $b\in\mathbb C$. This kind of integral appears often in quantum mechanics when computing the wave function using Fourier's transformation. However, it seem that physicists uses the formula $$\int_{-\infty}^\infty e^{-ax^2-bx}dx=\frac{1}{\sqrt a}\int_{-\infty}^\infty e^{-y^2+b^2/4a}dy=\frac{e^{b^2/4a}\sqrt\pi}{\sqrt a}$$ which I believe only holds when $a,b$ are real and $a>0$ unless there is further explanation.
Is there any way to justify this formula in the case of $a,b\in\mathbb C$ with ${\rm Re}(a)>0$? Or is there a workaround?
Update: what I feel confused about
My textbook and an answer below say "make the substitution $y=x+b/2a$". And the integral would be justified by $$I=\int_{-\infty}^{\infty} e^{-(ax^2+bx)} dx=\int_{-\infty}^{\infty} e^{-a(x^2+(b/a)x)} dx =\int_{-\infty}^{\infty} e^{-a([x+(b/(2a))]^2)} e^{b^2/(4a)} dx\\ =e^{b^2/(4a)} \int_{-\infty}^{\infty} e^{-ay^2} dy=\sqrt{\frac{\pi}{a}} e^{b^2/(4a)}$$
(1) After the substitution $y=x+b/2a$, $y$ will not be real if $b/2a$ is not. In this case, shouldn't the integral after the substitution be $$e^{b^2/(4a)} \int_{-\infty+b/2a}^{\infty+b/2a} e^{-ay^2} dy\quad ?$$ How is this defined? Perhaps a Riemann-Stieltjes integral?
(2) Suppose it is a Riemann-Stieltjes integral, this would mean we are integrating $e^{-ay^2}$, where $a$ and $y$ are both possibly complex, along the line ${\rm Im}(y)={\rm Im}(b/2a)$. How do I evaluate this?
(3) Suppose somehow we can indeed get rid of $b/2a$ and write $$\int_{-\infty}^\infty e^{-ax^2-bx}dx=e^{b^2/(4a)} \int_{-\infty}^{\infty} e^{-ay^2} dy$$ What's to justify $$\int_{-\infty}^{\infty} e^{-ay^2} dy=\sqrt{\pi/a}\quad?$$ when $a\notin\mathbb R$? I am assuming another substitution is made here: "$t=\sqrt{a}y$" to obtain $$\int_{-\infty}^{\infty} e^{-ay^2} dy=\sqrt{a}\int_L e^{-t^2} dt$$ where $L$ is the line $\mathbb R\cdot\sqrt a$. The well-known $\int_{-\infty}^\infty e^{-t^2}dt=\sqrt{\pi}$ does not seem to be usable here.
It is rather insightful to question whether the substitution is valid because the limits change in a potentially unpleasant way when the substitution is made. Let me show two ways to justify the substitution. First off, we could use contour integrals. Let $$-ax^2-bx=-a\left(x^2+\frac bax\right)=-a\left(x+\frac b{2a}\right)^2+\frac{b^2}{4a}=-a\left(x+f+ic\right)^2+\frac{b^2}{4a}$$ Where $f=\Re\frac b{2a}$ and $c=\Im \frac b{2a}$. Translation in the real direction doesn't change the limits but translation in the imaginary direction does: $$\int_{-\infty}^{\infty}e^{-ax^2-bx}dx=e^{\frac{b^2}{4a}}\int_{-\infty}^{\infty}e^{-a(x+f+ic)^2}dx=e^{\frac{b^2}{4a}}\int_{-\infty}^{\infty}e^{-a(y+ic)^2}dy$$ Where we have made the substitution $x+f=y$. To justify the imaginary translation we set up a contour integral $$\oint_Ca^{-az^2}dz=\int_{C_1}e^{-az^2}dz+\int_{C_2}e^{-az^2}dz+\int_{C_3}e^{-az^2}dz+\int_{C_4}e^{-az^2}dz$$ Where the contour $C$ is counterclockwise around a rectangle in the complex $z$-plane and the subcontours are shown in the following figure:
Now we can work out the integral over each subcontour. Along $C_1$, $z=t$ so $$\int_{C_1}e^{-az^2}dz=\int_{-N}^Ne^{-at^2}dt$$ So $$\lim_{t\rightarrow\infty}\int_{C_1}e^{-az^2}dz=\int_{-\infty}^{\infty}e^{-at^2}dt$$ Along $C_2$, $z=N+is$ so $$\int_{C_2}e^{-az^2}dz=i\int_0^ce^{-a(N+is)^2}ds$$ $$\begin{align}-a(N+is)^2&=-(\Re a+i\Im a)(N^2+2iNs-s^2)\\ &=-(N^2-s^2)\Re a+2Ns\Im a+i\left(-(N^2-s^2)\Im a-2Ns\Re a\right)\end{align}$$ So $$\left|e^{-a(N+is)^2}\right|=e^{-(N^2-s^2)\Re a+2Ns\Im a}$$ And $$\begin{align}\left|\int_{C_2}e^{-az^2}dz\right|&\le\int_0^ce^{-(N^2-s^2)\Re a+2Ns\Im a}ds\\ &\le\int_0^ce^{-(N^2-c^2)\Re a+2Nc\left|\Im a\right|}ds=ce^{-(N^2-c^2)\Re a+2Nc\left|\Im a\right|}\end{align}$$ So that $$\lim_{N\rightarrow\infty}\int_{C_2}e^{-az^2}dz=0$$ Along $C_3$, $z=t+ic$ so $$\int_{C_3}e^{-az^2}dz=\int_N^{-N}e^{-a(t+ic)^2}dt=-\int_{-N}^Ne^{-a(t+ic)^2}dt$$ So $$\lim_{N\rightarrow\infty}\int_{C_3}e^{-az^2}dz=-\int_{-\infty}^{\infty}e^{-a(t+ic)^2}dt$$ From the work surround the $C_2$ contour I hope it is clear that $$\left|\int_{C_4}e^{-az^2}dz\right|\le ce^{-(N^2-c^2)\Re a+2Nc\left|\Im a\right|}$$ So $$\lim_{N\rightarrow\infty}\int_{C_4}e^{-az^2}dz=0$$ Since the integrand is holomorphic along and within contour $C$ it follows that $$\oint_Ce^{-az^2}dz=0=\int_{-\infty}^{\infty}e^{-at^2}dt+0-\int_{-\infty}^{\infty}e^{-a(t+ic)^2}dt+0$$ So that justifies the imaginary translation step: $$\int_{-\infty}^{\infty}e^{-at^2}dt=\int_{-\infty}^{\infty}e^{-a(t+ic)^2}dt$$ Of course we only did it for $c=\Im\frac b{2a}>0$ you might have to repeat for $c<0$. So we are up to $$\int_{-\infty}^{\infty}e^{-ax^3-bx}dx=e^{\frac{b^2}{4a}}\int_{-\infty}^{\infty}e^{-ax^2}dx=2e^{\frac{b^2}{4a}}\int_0^{\infty}e^{-ax^2}dx$$ Now we let $a=re^{i\theta}$ and set up another contour integral
We want to do $$\oint_Ce^{-rz^2}dz=\int_{C_1}e^{-rz^2}dz+\int_{C_2}e^{-rz^2}dz+\int_{C_3}e^{-rz^2}dz$$ Along $C_1$, $z=t$ so $$\int_{C_1}e^{-rz^2}dz=\int_0^Re^{-rt^2}dt$$ and $$\lim_{R\rightarrow\infty}\int_{C_1}e^{-rz^2}dz=\int_0^{\infty}e^{-rt^2}dt=\frac1{\sqrt r}\int_0^{\infty}e^{-u^2}du=\frac12\sqrt{\frac{\pi}r}$$ Along $C_2$, $z=Re^{i\phi}$ so $$\int_{C_2}e^{-rz^2}dz=\int_0^{\frac{\theta}2}e^{-rR^2e^{2i\phi}}iRe^{i\phi}d\phi$$ Then $$-rR^2e^{2i\phi}=-rR^2\cos2\phi-irR^2\sin2\phi$$ So $$\left|e^{-rR^2e^{2i\phi}}\right|=e^{-rR^2\cos2\phi}\le e^{-rR^2\cos2\theta}$$ Thus $$\int_{C_2}e^{-rz^2}\le\int_0^{\frac{\theta}2}Re^{-rR^2\cos3\theta}d\phi=\frac{\theta}2Re^{-rR^2\cos3\theta}$$ And $$\lim_{R\rightarrow\infty}\int_{C_2}e^{-rz^2}dz=0$$ Along $C_3$, $z=te^{i\theta/2}$ so $$\int_{C_3}e^{-rz^2}dz=e^{\frac{i\theta}2}\int_R^0e^{-re^{i\theta}t^2}dt=-e^{\frac{i\theta}2}\int_0^Re^{-at^2}dt$$ So $$\lim_{R\rightarrow\infty}\int_{C_3}e^{-rz^2}dz=-e^{\frac{i\theta}2}\int_0^{\infty}e^{-at^2}dt$$ Again the integrand is holomorphic along and within contour $C$ so $$\oint_Ce^{-rz^2}dz=0=\frac1{\sqrt r}\int_0^{\infty}e^{-t^2}dt+0-e^{\frac{i\theta}2}\int_0^{\infty}e^{-at^2}dt$$ So now we have $$\int_{-\infty}^{\infty}e^{-ax^2-bx}dx=\frac{2e^{\frac{b^2}{4a}}}{\sqrt re^{\frac{i\theta}2}}\int_0^{\infty}e^{-t^2}dt=\frac{e^{\frac{b^2}{4a}}}{\sqrt a}\int_{-\infty}^{\infty}e^{-t^2}dt$$ Another way to justify is to start with $$I(c)=\int_{-\infty}^{\infty}e^{-a(y+ic)^2}dy$$ Then $$\frac{dI}{dc}=\int_{-\infty}^{\infty}-2ia(y+ic)e^{-a(y+ic)^2}dy=\left.ie^{-a(y+ic)^2}\right|_{-\infty}^{\infty}=0$$ So $$I(c)=\int_{-\infty}^{\infty}e^{-a(y+ic)^2}dy=\text{constant}=I(0)=\int_{-\infty}^{\infty}e^{-ay^2}dy$$ That gets us to $$\int_{-\infty}^{\infty}e^{-ax^3-bx}dx=2e^{\frac{b^2}{4a}}\int_0^{\infty}e^{-ax^2}dx$$ Now we can let $$J(\theta)=\int_0^{\infty}e^{-re^{i\theta}x^2}dx$$ Then $$\frac{dJ}{d\theta}=\int_0^{\infty}-rie^{i\theta}x^2e^{-re^{i\theta}x^2}dx=\left.\frac i2xe^{-re^{i\theta}x^2}\right|_0^{\infty}-\frac i2\int_0^{\infty}e^{-re^{i\theta}x^2}dx=-\frac i2J(\theta)$$ The general solution to this differential equation is $$J(\theta)=Ce^{-\frac i2\theta}$$ And we can apply the initial condition $$J(0)=C=\int_0^{\infty}e^{-rx^2}dx=\frac1{\sqrt r}\int_0^{\infty}e^{-t^2}dt$$ If we now let $a=re^{i\theta}$ then $$\int_{-\infty}^{\infty}e^{-ax^2-bx}dx=2e^{\frac{b^2}{4a}}J(\theta)=\frac{2e^{\frac{b^2}{4a}}}{\sqrt re^{\frac i2\theta}}\int_0^{\infty}e^{-t^2}dt=\frac{e^{\frac{b^2}{4a}}}{\sqrt a}\int_{-\infty}^{\infty}e^{-t^2}dt$$ So we could justify the imaginary translation step and the rotation step either via contour integrals or by differentiating with respect to a parameter.