# A improper integral with integrable singularities

Let $\alpha$ and $v_0$ be both positive. Consider a following integral: $${\mathcal J}^\alpha_{1/2}(v_0) := \int\limits_0^\infty v^{\alpha-1} e^{-v} \frac{1}{\sqrt{v_0-v}} d v$$ By splitting this integral into two integrals on the left and right hand sides of the singularity respectively, then in each integral by reforming the integrand appropriately and expanding a particular term into an infinite series and integrating term by term we have shown (and verified numerically ) that: $${\mathcal J}^\alpha_{1/2}(v_0) = \frac{\sqrt{\pi }}{ \cos (\pi a)} \left(e^{-i \pi a} v_0^{a-\frac{1}{2}} \Gamma (a) \, _1\tilde{F}_1\left(a;a+\frac{1}{2};-v_0\right)+i \sqrt{\pi } \, _1\tilde{F}_1\left(\frac{1}{2};\frac{3}{2}-a;-v_0\right)\right)$$ Here $_1\tilde{F}_1$ is the regularized hypergeometric function. In fact a much more general formula applies. $${\mathcal J}^\alpha_{n}(v_0) := \int\limits_0^\infty v^{\alpha-1} e^{-v} \frac{1}{(v_0-v)^n} d v$$ where $n \in (0,1)$. Then we have: $${\mathcal J}^\alpha_{n}(v_0) = \frac{\sin (\pi n) \Gamma (1-n)}{ \sin (\pi (a-n))} \cdot \\\left(e^{-i \pi n} \Gamma (n) \, _1\tilde{F}_1(n;-a+n+1;-v_0)-e^{-i \pi a} \Gamma (a) v_0^{a-n} \, _1\tilde{F}_1(a;a-n+1;-v_0)\right)$$ Now, the question is how does the result look like if the term in the square root in the denominator in the top first formula gets replaced by a polynomial in $v$ of some higher order?

• How $\sqrt{v_0-v}$ is defined when $v>v_0$? Commented Nov 2, 2015 at 18:57
• $\sqrt{v_0- v} = \imath \sqrt{v- v_0}$ for $v > v_0$. Commented Nov 2, 2015 at 19:00
• depends on the polynomial i would say Commented Nov 2, 2015 at 19:18

Take $\vec{v} := \left( v_j\right)_{j=0}^{d-1}$ where $0 < v_0 < v_1 < \cdots < v_{d-1} < \infty$ and define: $${\mathcal J}^\alpha_{1/2}\left( \vec{v} \right) := \int\limits_0^\infty v^{\alpha-1} e^{-v} \frac{1}{\sqrt{\prod\limits_{j=0}^{d-1} (v_j - v)}} dv$$ Even at the outset it is clear that the result will be expressed in terms of hypergeometric functions of many variables. Unfortunately systems like Mathematica or Matlab do not have those functions implemented in general. So even if we were able to derive the result in general it would have been hard to verify it numerically. For those reasons let us look at the case $d=2$ only. We have: \begin{eqnarray} {\mathcal J}^\alpha_{1/2}\left(v_0, v_1\right) = \int\limits_0^{v_0} \frac{v^{\alpha-1} e^{-v} dv}{\sqrt{\prod\limits_{j=0}^{1} (v_j - v)}} + \int\limits_{v_0}^{v_1} \frac{v^{\alpha-1} e^{-v} dv}{\sqrt{\prod\limits_{j=0}^{1} (v_j - v)}} + \int\limits_{v_1}^{\infty} \frac{v^{\alpha-1} e^{-v} dv}{\sqrt{\prod\limits_{j=0}^{1} (v_j - v)}} \end{eqnarray} We analyze the integrals from the left to the right. We have: \begin{eqnarray} 1^{st} &=& v_0^{\alpha-1/2} v_1^{-1/2} \int\limits_0^1 \frac{u^{\alpha-1} e^{-v_0 u} du}{\sqrt{(1-u)(1- \frac{v_0}{v_1} u)}}\\ &=& v_0^{\alpha-1/2} v_1^{-1/2} \frac{(\alpha-1)! (-1/2)!}{(\alpha-1/2)!}\sum\limits_{p=0}^\infty \sum\limits_{q=0}^\infty \frac{\alpha^{(p+q)}}{(\alpha+1/2)^{(p+q)}} \cdot (\frac{1}{2})^{(q)} \cdot \frac{(-v_0)^p}{p!} \cdot \frac{(v_0/v_1)^q}{q!} \end{eqnarray} The result can be expressed in terms of the Kampe de Ferriet function. The double sum is clearly convergent since $v_0/v_1 < 1$. \begin{eqnarray} 2^{nd} &=& -\imath v_1^{\alpha-1} e^{-v_1} \int\limits_0^1 \left( 1- \frac{v_1-v_0}{v_1} u \right)^{\alpha-1} e^{u(v_1-v_0)} \frac{d u}{\sqrt{u (1-u)}} \\ &=& -\imath v_1^{\alpha-1} e^{-v_1} ((-1/2)!)^2 \sum\limits_{p=0}^\infty \sum\limits_{q=0}^\infty \frac{(1/2)^{(p+q)}}{1^{(p+q)}} \cdot (1-\alpha)^{(q)} \cdot \frac{(v_1-v_0)^p}{p!} \cdot \frac{((v_1-v_0)/v_1)^q}{q!} \end{eqnarray} Now again the result is a Kampe de Ferriet function. The series are convergent for the same reasons as in the previous case. \begin{eqnarray} 3^{rd}&=& \int\limits_{v_1}^\infty u^{\alpha-2} e^{-u} \left(1-\frac{v_1}{u}\right)^{-1/2} \left(1-\frac{v_0}{u}\right)^{-1/2} d u \\ &=& (\alpha-2)! \sum\limits_{p=0}^\infty \sum\limits_{q=0}^\infty \frac{(1/2)^{(p)} (1/2)^{(q)}}{(2-\alpha)^{(p+q)}} \cdot \frac{(-v_1)^p}{p!} \cdot \frac{(-v_0)^q}{q!} + \\ &&\sqrt{\pi} v_1^{\alpha-1} \sum\limits_{p=0}^\infty \sum\limits_{q=0}^\infty \frac{(-\alpha-p+q)!}{(1/2-\alpha-p+q)!} \cdot (\frac{1}{2})^{(q)} \cdot (-v_1)^p \cdot (\frac{v_0}{v_1})^q \end{eqnarray} The later equation is obtained by expanding the two later factors in the integrand in binomial series and using the intergal representation of the upper incomplete Gamma function. That incomplete Gamma function is then split into a complete Gamma function ( which results in the first series) and a lower incomplete Gamma function. The later term is transformed by using the Taylor series expansion of the lower incomplete Gamma function and by swapping the order of summations. Both series are convergent due to the fact that the roots are ordered in ascending order. Below we show contour plots of the modulus of the integral as a function of $v_0$(X axis) and $v_1$(Y axis). Here we took $\alpha=1.2,1.6, 1.8$ from the left to the right respectively.
As we can see the modulus of the integral is a smooth bell shaped curve that attains maximum roughly at $v_0 = v_1 \simeq \alpha-1$.