I have to calculate integral: $$\int_{- \infty}^{\infty} \cfrac{\cos(z - \cfrac{1}{z})}{1 + z^2} dz$$ It is relay converge to $\cfrac{\pi}{e^2}$, but I have lost some opportunity to find solution.
EDIT: This was the OP's deleted attempt.
I have tried to reduce to: $$\int_{- \infty}^{\infty} \cfrac{\cos(z - \cfrac{1}{z})}{1 + z^2} dz= \Re \left( \int_{- \infty}^{\infty} \cfrac{\exp(i(x-1/x))}{1+x^2} dx \right) = $$ $$= \Re \left( \oint - \int_{0}^{\pi} \cfrac{\exp \left( i \left( \epsilon \exp(i \phi) - \cfrac{1}{\epsilon \exp(i \phi)} \right) \right)}{1+\epsilon^2 \exp(2 i \phi)}d\phi - \int_{\pi}^{0} \cfrac{\exp \left( i \left( R \exp(i \phi) - \cfrac{1}{R\exp(i \phi)} \right) \right)}{1+R^2 \exp(2 i \phi)}d\phi \right) = $$ $$\Re \left( - \int_{0}^{\pi} \cfrac{\exp \left( i \left( \epsilon \exp(i \phi) - \cfrac{1}{\epsilon \exp(i \phi)} \right) \right)}{1+\epsilon^2 \exp(2 i \phi)}d\phi + \int_{0}^{\pi} \cfrac{\exp \left( i \left( \cfrac{1}{\epsilon} \exp(i \phi) - \cfrac{\epsilon}{\exp(i \phi)} \right) \right)}{1+\cfrac{1}{\epsilon^2} \exp(2 i \phi)}d\phi \right)$$ or another way: $$\int_{- \infty}^{\infty} \cfrac{\cos(z - \cfrac{1}{z})}{1 + z^2} dz = \oint - \int_{0}^{\pi} \cfrac{\cos\left( i \left( \epsilon \exp(i \phi) - \cfrac{1}{\epsilon \exp(i \phi)} \right) \right)}{1+\epsilon^2 \exp(2 i \phi)}d\phi - \int_{0}^{\pi} \cfrac{\cos\left( i \left( R \exp(i \phi) - \cfrac{1}{R\exp(i \phi)} \right) \right)}{1+R^2 \exp(2 i \phi)} d\phi =$$ $$ \pi \cosh(2) - \int_{0}^{\pi} \cfrac{\cos\left( i \left( \epsilon \exp(i \phi) - \cfrac{1}{\epsilon \exp(i \phi)} \right) \right)}{1+\epsilon^2 \exp(2 i \phi)} d\phi$$ I don't exclude, that i have done a mistake, but it way i have tried.
