Conjectured Cauchy principal value for Goodwin-Staton integral

Question

In an answer to Goodwin Staton integral $G(x) = \int_0^\infty \frac{e^{-t^2}}{t+x}dt$ and its symmetry I conjectured $$PV \int_0^\infty \frac{e^{-t^2}}{t+x}dt = \sqrt{\pi} F(x) - \frac{1}{2} e^{-x^2}\mathrm{Ei}(x^2)\quad (x\ne 0)\quad(*)$$ where $F(x)$ is the Dawson integral and $\mathrm{Ei}(x)$ the exponential integral. This is an extension of http://dlmf.nist.gov/7.5.E13 to negative $x.$

Is there a rigorous proof for formula (*)?

I checked it numerically with the QUADPACK integration routine QAWC (see https://en.wikipedia.org/wiki/QUADPACK), which computes the Cauchy principal value of the integral of $f(x)/(x–c)$ for user-specified $c$ and $f$ and got the same results with $15$ digits:

   x            (*)                     QAWC
-10.00    -0.094123656234351    -0.094123656234351
 -9.00    -0.105340117792338    -0.105340117792338
 -8.00    -0.119603520771815    -0.119603520771815
 -7.00    -0.138358842389358    -0.138358842389358
 -6.00    -0.164146157367184    -0.164146157367184
 -5.00    -0.201901172247312    -0.201901172247312
 -4.00    -0.262774143418388    -0.262774143418388
 -3.00    -0.380019352972186    -0.380019352972186
 -2.00    -0.713887936713151    -0.713887936713151
 -1.00    -1.302308535738411    -1.302308535738411
  1.00     0.605133652503345     0.605133652503345
  2.00     0.354335928849531     0.354335928849531
  3.00     0.251934996448972     0.251934996448972
  4.00     0.195752582376989     0.195752582376989
  5.00     0.160154694796648     0.160154694796648
  6.00     0.135549871910491     0.135549871910491
  7.00     0.117516050600851     0.117516050600851
  8.00     0.103726368842216     0.103726368842216
  9.00     0.092838112910918     0.092838112910918
 10.00     0.084021593706602     0.084021593706602

Answer 1

Using this table, the Hilbert transform of $e^{-t^2}$ is $$\frac{1}{\pi} PV \int_{-\infty}^\infty \frac{e^{-t^2}}{x-t}dt = 2\pi^{-1/2} F(x).$$ Assume $x<0$. Then we have $$ 2\sqrt{\pi} F(x) = PV \int_{-\infty}^\infty \frac{e^{-t^2}}{x-t}dt = PV \int_{-\infty}^0 \frac{e^{-t^2}}{x-t}dt + PV \int_{0}^\infty \frac{e^{-t^2}}{x-t}dt=\\ PV \int_{-\infty}^0 \frac{e^{-t^2}}{x-t}dt - PV \int_{0}^\infty \frac{e^{-t^2}}{t-x}dt= PV \int_{-\infty}^0 \frac{e^{-t^2}}{x-t}dt - G(-x) $$ Substitution $t\rightarrow -t$ and reversing limits of integration gives $$ PV \int_{-\infty}^0 \frac{e^{-t^2}}{x-t}dt = -PV \int_{\infty}^0 \frac{e^{-t^2}}{x+t}dt = PV \int_0^{\infty}\frac{e^{-t^2}}{x+t}dt $$ and therefore $$ 2\sqrt{\pi} F(x) = PV \int_0^{\infty}\frac{e^{-t^2}}{x+t}dt - G(-x)$$

Then (*) follows from the fact that $F(x)$ is odd: $$PV \int_{0}^{\infty} \frac{e^{-t^2}}{x+t}dt =2\sqrt{\pi} F(x) +G(-x) = 2\sqrt{\pi} F(x) + \sqrt{\pi} F(-x)- \frac{1}{2} e^{-x^2}\mathrm{Ei}(x^2)\\ =\sqrt{\pi} F(x) - \frac{1}{2} e^{-x^2}\mathrm{Ei}(x^2) $$