$\int_{-\infty}^{\infty}x e^{-x^2 + x(i+1)}dx$ how to?

Question

I am having some problems in solving the integral reported in the title, which is:

$$\int_{-\infty}^{\infty}x e^{-x^2 + x(i+1)}dx$$

As from the general theory of Gaussian integrals I have been trying to write the known integral

$$\int_{-\infty}^{\infty} e^{-x^2 - i k x}dx, \; k \in \mathbb{R}$$

which has a solution with contour integration knowing that the complex function $e^{-z^2}$ is holomorphic on the whole complex plane. Despite my efforts this route does not seem to bring to any result.

Thanks to anyone who is so keen to try and help me.

Answer 1

Hint: compute $\int_{-\infty}^{\infty} xe^{-x^{2}+tx} dx$ for real $t$ by writing $-x^{2}+tx$ as $-(x-\frac t 2)^{2}+\frac {t^{2}} 4$. You will get a nice answer $f(t)$ in which you can replace $t$ by a complex number $z$ to get an entire function. The equation $f(z)=\int_{-\infty}^{\infty} xe^{-x^{2}+zx} dx$ now holds for all complex $z$ because if two entire functions coincide on the real line they coincide everywhere. Now put $z=1+i$.

Answer 2

First note that $\int_{-\infty}^\infty(-2x+i+1)e^{-x^2+x(i+1)}dx=[e^{-x^2+x(i+1)}]_{-\infty}^\infty=0$, so your integral is $$\frac{i+1}{2}\int_{\Bbb R}e^{-x^2+(i+1)x}dx.$$But $$\int_{\Bbb R}e^{-x^2+zx}dx=\int_{\Bbb R}e^{-y^2+z^2/4}dy=\sqrt{\pi}e^{z^2/4},$$so your final answer is $$\frac{i+1}{2}\sqrt{\pi}e^{(1+i)^2/4}=\sqrt{\frac{\pi}{2}}e^{i(1/2+\pi/4)}.$$

Answer 3

We can treat the integral as the $\xi = 1$ value of the Fourier transform integral $$ \int_{-\infty}^\infty e^{ix\xi}xe^{-x^2 + x}~dx. $$ First, we can turn the exponential factor into a true Gaussian by completing the square: $$ x^2 - x = (x - \frac{1}{2})^2 - \frac{1}{4}, $$ and hence the integral is given by $$ e^{\frac{1}{4}} \int_{-\infty}^\infty e^{ix\xi}xe^{-(x-\frac{1}{2})^2}~dx. $$ Next we make the change of variable $y = x-\frac{1}{2}$, after which the integral becomes (ignoring the prefactor) $$ \int_{-\infty}^\infty e^{iy\xi} (y+\frac{1}{2}) e^{-y^2}~dy = e^{i\frac{1}{2}\xi}\int_{-\infty}^\infty e^{iy\xi}ye^{-y^2}~dy + \frac{e^{i\frac{1}{2}\xi}}{2}\int_{-\infty}^\infty e^{iy\xi}e^{-y^2}~dy. $$ Each of the latter two integrals can be done individually. The second is simply the Fourier transform of a Gaussian. The first can be reduced to computing the Fourier transform of a Gaussian, either by noting that $$ \int_{-\infty}^\infty e^{iy\xi}ye^{-y^2}~dy = \frac{1}{i}\frac{d}{d\xi} \int_{-\infty}^\infty e^{iy\xi}e^{-y^2}~dy $$ or by integrating by parts, noting that $$ ye^{-y^2} = -\frac{1}{2}\frac{d}{dy}e^{-y^2}. $$

Answer 4

Hint:

$$\int_{x=-\infty}^\infty xe^{-x^2+(1+i)x}dx\\ =\int_{x=-\infty}^\infty\left(x-\frac{1+i}2\right)e^{((1+i)/2)^2-(x-(1+i)/2)^2}dx +\frac{1+i}2\int_{x=-\infty}^\infty e^{((1+i)/2)^2-(x-(1+i)/2)^2}dx\\ =e^{i/2}\int_{x=-\infty}^\infty ze^{-z^2}dz +\frac{1+i}2e^{i/2}\int_{x=-\infty}^\infty e^{-z^2}dz.$$

The first integral vanishes by parity and the second is a more classical Gaussian integral.