integral with branch cut (depending on parameter)

Question

I am trying to work out the integral \begin{equation} \int_{-1}^{+1} dx \frac{1+i \lambda x}{2(1-\lambda^2+2i \lambda x)^\frac{3}{2}}, \end{equation} where $\lambda$ is real number and $i$ the imaginary unit. I am struggling with choosing an integral contour since the branch cut depends on $\lambda$. By numerical evaluations, I suspect for $|\lambda|<1$, the integral is 1, otherwise it is zero. Is there a proper way to do this integral?

Answer 1

Hint First write your integrand as :

$$ \frac{1+i \lambda x}{2(1-\lambda^2+2i \lambda x)^\frac{3}{2}} =\frac{2 + \lambda^2+2i \lambda\color{red}{(x+\frac{i\lambda}{2}) }}{\color{blue}{4}(1+2i \lambda \color{red}{(x+\frac{i\lambda}{2}) })^\frac{3}{2}} \\:=\frac{2 + \lambda^2+2i \lambda\color{red}{z }}{\color{blue}{4}(1+2i \lambda \color{red}{z })^\frac{3}{2}} =f_\lambda(z) $$

Where, e set $\color{blue}{z =x+\frac{i\lambda}{2}},$ that is $dx =dz$. Then, with the change of variables $\color{blue}{z =x+\frac{i\lambda}{2}},$ we get, $$I:=\int_{-1}^{+1} \frac{1+i \lambda x}{2(1-\lambda^2+2i \lambda x)^\frac{3}{2}} dx =\int_{\color{blue}{(-1, \frac{\lambda}{2})}}^{\color{blue}{(1, \frac{\lambda}{2})}} f_\lambda(z) dz $$

From it is easy since can use the contour integral on the rectangle $ R_\lambda$with vertices's $(-1,0)-(1,0)-\color{blue}{(1, \frac{\lambda}{2})}-\color{blue}{(-1, \frac{\lambda}{2})}$

Remark This will very helpful, $f_\lambda$ has one pole which is $z_\lambda = \frac{i\lambda}{2}$ and we have,

$$\color{brown}{\oint_{R_\lambda} f_\lambda(z) dz }= \color{blue}{\int_{-1}^{+1}f_\lambda(z) dz} +\int_{\color{blue}{(1,0)}}^{\color{blue}{(1, \frac{\lambda}{2})}} f_\lambda(z) dz +\int^{\color{blue}{(-1, \frac{\lambda}{2})}}_{\color{blue}{(1, \frac{\lambda}{2})}} f_\lambda(z) dz\\+\int^{\color{blue}{(-1,0)}}_{\color{blue}{(-1, \frac{\lambda}{2})}} f_\lambda(z) dz \\=-I+ \color{blue}{\int_{-1}^{+1}}\frac{2 + \lambda^2+2i \lambda\color{red}{x }}{\color{blue}{4}(1+2i \lambda \color{red}{x})^\frac{3}{2}}dx+ \int_{\color{blue}{(1,0)}}^{\color{blue}{(1, \frac{\lambda}{2})}} f_\lambda(z) dz+\int^{\color{blue}{(-1,0)}}_{\color{blue}{(-1, \frac{\lambda}{2})}}f_\lambda(z) dz $$