# Solve analytically an equation with a definite integral

Given a probability distribution $$p(u)$$, I need to solve the following system where $$\gamma,\; \alpha \in \mathbb{R}$$ are the unknowns: \begin{align} 1=\int_{-\infty}^{\infty} p(u) \frac{-1}{\gamma-(u-x+\alpha \tau)^{2}}\mathrm{d}u \end{align} \begin{align} \alpha=\int_{-\infty}^{\infty} p(u) \frac{(u-x+\alpha \tau)}{\gamma-(u-x+\alpha \tau)^{2}} \end{align}

I am only interested in an expression for $$\alpha$$, and I have been solving this system numerically.

Is there a way to simplify these equations for a general $$p(u)$$ ?

If not, if we choose $$p(u)$$ to be a simple distribution (bimodal, uniform for example), is is then possible to simplify these equations without using numerical methods ?

($$x$$ is a real number, and $$\tau$$ is a parameter such that $$-1<\tau<1$$ and $$\tau \neq 0$$).

Any remark, advice or reference to a book is always appreciated, thank you very much.

Edit :

The original equation has a dependency on an extra term $$y\in\mathbb{R}$$ :

\begin{align} 1=\int_{-\infty}^{\infty} p(u) \frac{-1}{\gamma-(u-x+\alpha \tau)^{2}-\frac{y^2}{\left(\tau-1\right)^{2}}}\mathrm{d}u \end{align} \begin{align} \alpha=\int_{-\infty}^{\infty} p(u) \frac{(u-x+\alpha \tau)}{\gamma-(u-x+\alpha \tau)^{2}-\frac{y^2}{\left(\tau-1\right)^{2}}} \end{align} I noticed that for any distribution $$p(u)$$ my solution is independent of $$y$$ (I don't know why) so I thought that setting $$y=0$$ would make things easier.

• How do you handle the singularity when the denominator vanishes? Aug 24 '20 at 15:34
• I edited the question. There is an extra term in the original equation, so the denominator should never vanish I think.
– Matt
Aug 24 '20 at 16:55
• The denominator still vanishes, so presumably you're calculating a principle value integral. Aug 24 '20 at 17:32
• I am not sure I understand, could you explain?
– Matt
Aug 24 '20 at 18:15

Let us consider the given integral equations in the form of

\left\{\begin{align} &1 = \int\limits_{-\infty}^\infty \dfrac{1}{(u-a)^2 - b^2}\,p(u)\text{ d}u\\ &\alpha = \int\limits_{-\infty}^\infty \dfrac{a-u}{(u-a)^2-b^2}\,p(u)\text{ d}u, \end{align}\right.\tag1 where $$a=x-\alpha\tau,\quad b^2 = \gamma-\dfrac {y^2}{(r-1)^2}.\tag2$$

Then \left\{\begin{align} &b+\alpha = \int\limits_{-\infty}^\infty \dfrac{b+a-u}{(u-a)^2 - b^2}\,p(u)\text{ d}u = \int\limits_{-\infty}^\infty p(u)\dfrac{\text{ d}u}{a-u-b}\\ &b-\alpha = \int\limits_{-\infty}^\infty \dfrac{b+u-a}{(u-a)^2-b^2}\,p(u)\text{ d}u = \int\limits_{-\infty}^\infty p(u)\dfrac{\text{ d}u}{u-a-b}, \end{align}\right.\tag3

Obtained integrals allow more rich choice of the further steps. Detalization of the function $$p(u)$$ looks nesessary.

$$\color{green}{\mathbf{Case\ p(u) = \frac12\delta(u+1)-\frac12\delta(u-1)}}.$$

Via (1).

Formulas $$(1)$$ give the system

\left\{\begin{align} &1 = \frac12\dfrac1{(u-a)^2 - b^2}\bigg|_{-1}^1 = \frac12\dfrac1{(a-1)^2 - b^2}-\frac12\dfrac1{(a+1)^2 - b^2}\\ &\alpha = \frac12\dfrac{a-u}{(u-a)^2-b^2}\bigg|_{-1}^1 = \frac12\dfrac{a-1}{(a-1)^2-b^2}-\frac12\dfrac{a+1}{(a+1)^2-b^2}, \end{align}\right.

$$\begin{cases} v=u-2\\[4pt] (a-1)u-(a+1)v = 2\alpha\\ u = \dfrac1{(a-1)^2 - b^2}\\ v = \dfrac{a+1}{(a+1)^2-b^2} \end{cases}\Rightarrow \begin{cases} u = a-\alpha+1\\[4pt] v = a-\alpha-1\\ a-\alpha+1 = \dfrac1{(a-1)^2 - b^2}\\ a-\alpha-1 = \dfrac{1}{(a+1)^2-b^2}, \end{cases}$$ $$\begin{cases} (a-1)^2 - b^2 = \dfrac1{a-\alpha+1}\\ (a+1)^2 - b^2 = \dfrac1{a-\alpha -1}, \end{cases}$$

$$2a(a-\alpha)^2-2a-1 = 0,\tag4$$ i.e. $$\alpha$$ does not depend of $$b.$$

Let $$p=\dfrac\tau{\tau+1},\quad q=\dfrac x{\tau+1},\quad z= a-\alpha = x-(\tau+1)\alpha,\tag5$$ then $$a=z+\alpha = z + \frac{x-z}{\tau+1} = \dfrac x{\tau+1}+ \dfrac\tau{\tau+1}z = p z + q,$$ $$\alpha = a-z = (p-1)$$ and from $$(4)$$ should $$2(pz+q)(z^2-1)-1=0,$$ with the solution $$z = \frac1{6p}\left(r - \frac{12 p^2 + 4 q^2}r + 2 q\right),\tag5$$ where $$r = \sqrt[3]{\sqrt{(-72 p^2 q - 54 p^2 + 8 q^3)^2 - (12 p^2 + 4 q^2)^3} - 72 p^2 q - 54 p^2 + 8 q^3}.\tag6$$

Note that $$(5)$$ gives correct solution even if $$p$$ is complex.

Via (3).

Formulas $$(3)$$ give the system

\left\{\begin{align} &b+\alpha = \frac12\dfrac1{a-u-b}\bigg|_{-1}^1 = \frac12\dfrac1{a-1-b}-\frac12\dfrac1{a+1-b} = \dfrac 1{(a-b)^2-1}\\ &b-\alpha = \frac12\dfrac1{u-a-b}\bigg|_{-1}^1 = \frac12\dfrac1{1-a-b}+\frac12\dfrac1{1+a+b} = \dfrac1{1-(a+b)^2}, \end{align}\right.

$$\begin{cases} (\alpha+b)(a^2-2ab+b^2-1) = 1\\[4pt] (\alpha-b)(a^2+2ab+b^2-1) = 1 \end{cases}\Rightarrow \begin{cases} \alpha(-2ab)+b(a^2+b^2-1) = 0\\[4pt] \alpha(a^2+b^2-1)+b(-2ab) = 1, \end{cases}$$$$\begin{cases} (\alpha^2-b^2)(-2ab) = -b\\ (\alpha^2-b^2)(a^2+b^2-1) = \alpha\\ \end{cases} \begin{cases} 2a(\alpha^2-b^2) = 1\\ a^2+b^2-1 = 2\alpha a \end{cases}$$

$$2a(\alpha^2+a^2 - 2\alpha a -1) =1$$ with the same equation $$(4).$$

• Let us say $p(u)=\frac{1}{2}\delta(u+1)+\frac{1}{2}\delta(u-1)$, would it be possible to have an analytic solution for $\alpha$? It seems that solving numerically still is the best way to go...
– Matt
Aug 27 '20 at 21:38
• @Matt Thank you! The case is ready to check. If we have the complex $r,$ then the real solution via trig functions exists. Aug 29 '20 at 4:48