show that $$\int_{-\infty}^{+\infty}\frac{\Re(1+(x-ib)^4)}{\left | (1+(x+ib)^4 \right |^2}dx=\int_{-\infty}^{+\infty}\frac{dx}{1+x^4}:0<b<\frac{1}{\sqrt 2}$$
I think to exapand $(x+ib)^4$ and $(x-ib)^4$ but Iam sure who put the question at the book has another way so I dont try with it .
also I think with $(x+ib)=Re^{i\theta}$so $(x+ib)=Re^{-i\theta}$ but the integral became very complicated for me .
another thing :$\left | (1+(x+ib)^4 \right |=\left | (1+(x-ib)^4 \right |$so $$\int_{-\infty}^{+\infty}\frac{\Re(1+(x-ib)^4)}{\left | (1+(x+ib)^4 \right |^2}dx=\int_{-\infty}^{+\infty}\frac{\Re(1+(x-ib)^4)}{\left | (1+(x-ib)^4 \right |^2}dx=\int_{-\infty}^{+\infty}\frac{\Re(1+(x-ib)^4)}{\Re^2(1+(x-ib)^4)+\Im^2(1+(x-ib)^4)}dx$$
but also I faild to find way to slove it
so can any one give an advice which make me able to solve it ? and if I faild to solve it with the advice which I take it from here , iam going to ask who give me the advice about full answer
thanks for all
Pull the real part out of the integral,
$$\int_{-\infty}^\infty \frac{\Re (1 + (x-ib)^4)}{\lvert 1 + (x+ib)^4\rvert^2}\, dx = \Re \int_{-\infty}^\infty \frac{1 + (x-ib)^4}{\lvert 1 + (x+ib)^4\rvert^2}\, dx$$
Now you can cancel the $1+(x-ib)^4$ to get
$$\Re \int_{-\infty}^\infty \frac{dx}{1 + (x+ib)^4} = \Re \int_{-\infty+ib}^{\infty+ib} \frac{dz}{1+z^4}.$$
The integrand has no poles between the real line and the line $\Im z = b$, since $0 < b < 1/\sqrt{2}$. Thus for every $R > 2$, the integral over the boundary of the rectangle with vertices at $-R$, $R$, $R+ib$, $-R+ib$ is $0$ by Cauchy's integral theorem. For $\Re z = \pm R$, the integrand is bounded by $2/R^4$, so the integral over the vertical sides of the rectangle tends to $0$, and letting $R\to \infty$, we obtain
$$\Re \int_{-\infty+ib}^{\infty+ib} \frac{dz}{1+z^4} = \Re \int_{-\infty}^\infty \frac{dx}{1+x^4},$$
but of course the last integral is real, so we can also drop the $\Re$.