# Conjectured Cauchy principal value for Goodwin-Staton integral

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

• This is the Mellin convolution of $\exp(-t^2)$ and $\frac1{1+t}$. Thus, you can take the Mellin transforms of these two, multiply those and express fully as gamma functions, and then invert the transform. – J. M. is a poor mathematician Aug 14 '17 at 19:15
• @j-m-isnt-a-mathematician: I am not familiar with Mellin transforms. Using your comment, the table from mathworld.wolfram.com/MellinTransform.html, and the Gamma reflection formula I get $$MT\left(e^{-t^2}\right)MT\left(\frac{1}{1+t}\right)=\pi \csc\left(\pi z\right)\times \frac{1}{2} \Gamma\left(\frac{1}{2}z\right)=\frac{1}{2}\Gamma\left(\frac{1}{2}z\right) \Gamma\left(z\right)\Gamma\left(1-z\right).$$ But how do I get the inverse transform and how does this prove the PV formula? – gammatester Aug 14 '17 at 20:02
• After converting the $\Gamma\left(\frac{z}{2}\right)$, you should now be able to get a corresponding Meijer $G$ expression that should reduce to a hypergeometric case, since the functions in the closed form are hypergeometric. Anyway, that was only an idea; there is possibly a simpler route. – J. M. is a poor mathematician Aug 14 '17 at 20:49
• @j-m-is-not-a-mathematician: I think I found another route, would you please have a look on my answer. Are there irreparable errors? – gammatester Aug 16 '17 at 11:50
• You will like the question (and hopefully the answers) here: math.stackexchange.com/questions/1412761/… – tired Aug 17 '17 at 21:40

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: $$PV \int_{0}^{\infty} \frac{e^{-t^2}}{x+t}dt =2\sqrt{\pi} F(x) +G(-x) = 2\sqrt{\pi} F(x) + \sqrt{\pi} F(-x)- \frac{1}{2} e^{-x^2}\mathrm{Ei}(x^2)\\ =\sqrt{\pi} F(x) - \frac{1}{2} e^{-x^2}\mathrm{Ei}(x^2)$$

• I didn't have a Hilbert transform table on hand, but this looks quite clean! – J. M. is a poor mathematician Aug 16 '17 at 14:20
• @j-m-is-not-a-mathematician: Many thanks for your effort! – gammatester Aug 16 '17 at 14:40