Double gaussian integral with a real pole

Question

How to solve the following double integral with a real pole? $$\int_{-\infty}^{+\infty}dxdy\, \frac{x}{y(y-x)}\, e^{-\alpha^2 (x^2+y^2)+i\beta\, (y-x)}$$ I am able to solve it by first deriving with respect to $\beta$ (in order to cancel out the pole) and then integrating back in $\beta$; but doing this way there is an integration constant left, which I don't know how to compute.

Answer 1

$$I(\alpha, \beta)=\int_{-\infty}^\infty\int_{-\infty}^\infty dxdy\, \frac{x}{y(y-x)}\, e^{-\alpha^2 (x^2+y^2)+i\beta\, (y-x)}$$ $$=-\int_{-\infty}^\infty\int_{-\infty}^\infty dxdy\Big(\frac{1}{y}+\frac{1}{x-y} \Big) e^{-\alpha^2 (x^2+y^2)+i\beta\, (y-x)}=I_1+I_2$$ Using the symmetry of the integrand, $$I_1=-i\int_{-\infty}^\infty e^{-\alpha^2 x^2-i\beta x}dx\int_{-\infty}^\infty\frac{\sin\beta y}{y}e^{-\alpha^2 y^2}dy$$ To evaluate $\displaystyle \int_{-\infty}^\infty\frac{\sin\beta y}{t}e^{-\alpha^2 y^2}dy=\int_{-\infty}^\infty\frac{\sin t}{t}e^{-a^2 t^2}dt;\, a=\frac{\alpha}{\beta}$, we can use the Parseval' theorem $$\int_{-\infty}^\infty f(x)g(x)dx=\frac{1}{2\pi}\int_{-\infty}^\infty \hat f(k)\hat g^*(k)dk,\,\,\text{where} \,\, \hat f(k)=\int_{-\infty}^\infty f(x)e^{ikx}dx$$ Making a full square, $$\int_{-\infty}^\infty e^{-a^2 t^2}e^{ikt}dt=\int_{-\infty}^\infty e^{-a^2\big( t^2+\frac{ik t}{a^2}-\frac{k^2}{4a^4}+\frac{k^2}{4a^4}\Big)}dt=e^{-\frac{k^2}{4a^2}}\int_{-\infty}^\infty e^{-a^2\big(t-\frac{ik}{2a^2}\big)^2}dt=\frac{\sqrt \pi}{a}e^{-\frac{k^2}{4a^2}}$$ $$\int_{-\infty}^\infty \frac{\sin t}{t}e^{ikt}dt=\frac{1}{2i}\int_{-\infty}^\infty \frac{e^{it(k+1)}-e^{it(k-1)}}{t}dt$$ Using the symmetry of the integral and making changes of the variable $x=t(k\pm1)$, it is not difficult to show that the integral is not equal to zero only at $k\in[-1;1]$, and for $k$ lying in this interval we get: $$\frac{1}{2i}\int_{-\infty}^\infty \frac{e^{it(k+1)}-e^{it(k-1)}}{t}dt=\frac{1}{2i}\int_{-\infty}^\infty \frac{2i\sin t}{t}dt=\pi; \,\,k\in[-1;1]$$ Therefore, using the Parseval' theorem $$\int_{-\infty}^\infty\frac{\sin\beta y}{y}e^{-\alpha^2 y^2}dy=\int_{-\infty}^\infty\frac{\sin t}{t}e^{-a^2 t^2}dt=\frac{1}{2\pi}\frac{\sqrt\pi}{a}\pi\int_{-1}^1e^{-\frac{k^2}{4a^2}}dk$$ $$=2\sqrt\pi\int_0^\frac{1}{2a}e^{-t^2}dt=\pi\operatorname{erf}\Big(\frac{1}{2a}\Big)=\pi\operatorname{erf}\Big(\frac{\beta}{2\alpha}\Big)\qquad(1)$$ Making a full square again, we find that $$\int_{-\infty}^\infty e^{-\alpha^2 x^2-i\beta x}dx=\frac{\sqrt\pi}{\alpha}e^{-\frac{\beta^2}{4\alpha^2}}$$ and $$I_1=-i\pi\frac{\sqrt\pi}{\alpha}e^{-\frac{\beta^2}{4\alpha^2}}\operatorname{erf}\Big(\frac{\beta}{2\alpha}\Big)\qquad (2)$$ Now, making the substitution $t=x-y$ and $s=x+y$, using $x^2+y^2=\frac{1}{2}(t^2+s^2)$, and evaluating the Jacobian ($=\frac{1}{2}$), we get for $I_2$ $$I_2=-\frac{1}{2}\int_{-\infty}^\infty e^{-\frac{\alpha^2}{2} s^2}ds\int_{-\infty}^\infty e^{-\frac{\alpha^2}{2} t^2-i\beta t}\frac{dt}{t}=\frac{i}{2}\int_{-\infty}^\infty e^{-\frac{\alpha^2}{2} s^2}ds\int_{-\infty}^\infty e^{-\frac{\alpha^2}{2} t^2}\frac{\sin \beta t}{t}dt$$ Using (1) (we just have to change $\alpha\to\frac{\alpha}{\sqrt2}$) $$I_2=\frac{i\pi}{2}\frac{\sqrt{2\pi}}{\alpha}\operatorname{erf}\Big(\frac{\beta}{\sqrt2\alpha}\Big)\qquad(3)$$ Using (2) and (3) $$\boxed{\,\,I(\alpha, \beta)=I_1+I_2=-i\pi\frac{\sqrt\pi}{\alpha}\bigg(e^{-\frac{\beta^2}{4\alpha^2}}\operatorname{erf}\Big(\frac{\beta}{2\alpha}\Big)-\frac{1}{\sqrt2}\operatorname{erf}\Big(\frac{\beta}{\sqrt2\alpha}\Big)\bigg)\,\,}\qquad(4)$$

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To check the answer, we can consider the limit $\beta\to 0$.

Decomposing the exponent in the powers of $\beta$ $$I(\alpha, \beta)=\int_{-\infty}^\infty\int_{-\infty}^\infty \frac{xe^{-\alpha^2 (x^2+y^2)}}{y(y-x)} \Big(1+i\beta(y-x)-\frac{\beta^2}{2}(y-x)^2-\frac{i\beta^2}{2}(y-x)^3+..\Big)dxdy $$ $$=\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\alpha^2 (x^2+y^2)}\Big(\frac{x}{y(y-x)}+i\beta\frac{x}{y}-\frac{\beta^2}{2}\big(x-\frac{x^2}{y}\big)-\frac{i\beta^3}{6}\frac{x(y-x)^2}{y}+o(\beta^3)\Big)dxdy$$ It is easy to see that due to the symmetry of the integral "survive" only even terms (with regard to both $x$ and $y$), and the biggest such term is $\sim O(\beta^3)$: $$I(\alpha, \beta)=\frac{2i\beta^3}{6}\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-\alpha^2 (x^2+y^2)}x^2dxdy+o(\beta^3)=\frac{i\pi}{6}\frac{\beta^3}{\alpha^4}+o(\beta^3)$$ On the other hand, decomposing (4) $$I(\alpha, \beta)=-i\pi\frac{\sqrt\pi}{\alpha}\frac{2}{\sqrt\pi}\bigg(e^{-\frac{\beta^2}{4\alpha^2}}\int_0^\frac{\beta}{2\alpha}e^{-t^2}dt-\frac{1}{\sqrt2}\int_0^\frac{\beta}{\sqrt2\alpha}e^{-t^2}dt\bigg)$$ $$=-\frac{2\pi i}{\alpha}\bigg(\big(1-\frac{\beta^2} {4\alpha^2}\big)\int_0^\frac{\beta}{2\alpha}(1-t^2)dt-\frac{1}{\sqrt 2}\int_0^\frac{\beta}{\sqrt2\alpha}(1-t^2)dt\bigg)+...$$ $$=-\frac{2\pi i}{\alpha}\bigg(\big(1-\frac{\beta^2} {4\alpha^2}\big)\Big(\frac{\beta}{2\alpha}-\frac{1}{3}\big(\frac{\beta}{2\alpha}\big)^3\Big)-\frac{1}{\sqrt2}\Big(\frac{\beta}{\sqrt2\alpha}-\frac{1}{3}\big(\frac{\beta}{\sqrt2\alpha}\big)^3\Big)\bigg)+...$$ $$=-\frac{2\pi i}{\alpha}\frac{\beta^3}{4\alpha^3}\Big(-\frac{1}{6}-\frac{1}{2}+\frac{1}{3}\Big)+O(\beta^5)=\frac{i\pi}{6}\frac{\beta^3}{\alpha^4}+O(\beta^5)$$