I am a little new in complex analysis, so be patient with me:

First, I have defined a function $ f (x) = frac { sin (x- frac 1x)} {x + frac 1x} $, so that I can define a new function $ f (z) = frac {e ^ {i (z- frac 1z)}} {z + frac 1z} $, which I then simplified to be $ f (z) = frac {ze ^ {i (z- frac 1z)}} {z ^ 2 + 1} $

Second, I defined an outline $ C $ which contains the interval $ (- R, – epsilon) $, contour $ gamma $ who has a half circle of radius $ epsilon $ centered at the origin, the interval $ ( epsilon, R) $and the semicircular outline $ Gamma $ with radius $ R $ centered at the origin. I want to take the limit as $ R rightarrow infty $ and $ epsilon rightarrow0 $

So,

$$ int_C f (z) dz = int _ {- R} ^ {- epsilon} f (z) dz + int_ gamma f (z) dz + int_ epsilon ^ R f (z) dz + int_ Gamma f (z) dz $$

For the leftmost integral, I can simply use the residue theorem, as $ C $ is a closed outline with a single pole to $ i $. So,

$$ int_C f (z) dz = 2 pi i Res (f, i) $$

$$ = 2 pi i frac {1} {2e ^ 2} = frac { pi i} {e ^ 2} $$

Then, looking at the first and third integrals on the right,

replace $ z = -u $ and $ dz = -du $ in the first integral, then

$$ int _ {- R} ^ {- epsilon} f (z) dz + int_ epsilon ^ R f (z) dz $$

$$ int _ { epsilon} ^ {R} frac {-ue ^ {i ( frac 1u-u)}} {u ^ 2 + 1} of + int # epsilon ^ R frac {ze ^ {i (z- frac 1z)}} {z ^ 2 + 1} dz $$

which can then be combined to get

$$ int _ { epsilon} ^ {R} frac {z (e ^ {i (z- frac 1z)} – e ^ {- i (z- frac 1z)})} {z ^ 2 + 1} dz $$

Then, multiplying by $ frac {2i} {2i} $, we have

$$ 2i int _ { epsilon} ^ {R} enac {z sin (z- frac 1z)} {z ^ 2 + 1} dz $$

And then, leaving $ R rightarrow infty $ and $ epsilon rightarrow0 $ we have

$$ 2i int_ {0} ^ { infty} enac {z sin (z- frac 1z)} {z ^ 2 + 1} dz $$

Moreover, since this integral is only on the real line, we can exchange the $ z $ for a $ x $, and also simplify for it to look like the original

$$ 2i int_ {0} ^ { infty} enac { sin (x- frac 1x)} {x + frac 1x} dx $$

who then leaves us with

$$ frac { pi i} {e ^ 2} = 2i int_ {0} ^ { infty} frac { sin (x- frac 1x)} {x + frac 1x} dx + int_ gamma f (z) dz + int_ Gamma f (z) dz $$

The problem is that I do not really know how to deal with the other two integral parts of the equation. I'm sure the integral on $ Gamma $ tends to 0 simply using the M-L inequality, but I'm not so sure of how to evaluate the integral on $ gamma $