Poiché l'integrale che hai proposto, dovendo esprimersi mediante la seguente funzione speciale:
Exponential integral
$Ei(x)=-\int_{-x}^{+oo}e^(-t)/tdt$
è tutt'altro che banale:
Wolfram
$\int_{0}^{+oo}sinx/(x^2+1)dx=(Ei(1)-e^2Ei(-1))/(2e)~~0.646761$
sorge il dubbio che sia il frutto di una tua iniziativa personale. Ad ogni modo, premesso che:
$\int_{0}^{+oo}sinx/(x^2+1)dx=$
$=1/(2i)\int_{0}^{+oo}e^(ix)/(x^2+1)dx-1/(2i)\int_{0}^{+oo}e^(-ix)/(x^2+1)dx=$
$=-1/4\int_{0}^{+oo}e^(ix)/(x-i)dx+1/4\int_{0}^{+oo}e^(ix)/(x+i)dx+1/4\int_{0}^{+oo}e^(-ix)/(x-i)dx-1/4\int_{0}^{+oo}e^(-ix)/(x+i)dx$
si può procedere considerando le seguenti 4 funzioni ausiliarie e i loro rispettivi percorsi d'integrazione.
1. $f_1(z)=e^(iz)/(z-i)$
$\gamma_1$ percorsa in senso antiorario
$\int_{\gamma_1}e^(iz)/(z-i)dz=0 rarr$
$rarr \int_{0}^{R}e^(ix)/(x-i)dx-i\int_{0}^{1-r}e^(-t)/(it-i)dt-i\int_{1+r}^{R}e^(-t)/(it-i)dt+\int_{C_R}e^(iz)/(z-i)dz+\int_{C_r}e^(iz)/(z-i)dz=0 rarr$
$rarr \int_{0}^{R}e^(ix)/(x-i)dx-1/e\int_{0}^{1-r}e^(-(t-1))/(t-1)dt-1/e\int_{1+r}^{R}e^(-(t-1))/(t-1)dt+\int_{C_R}e^(iz)/(z-i)dz+\int_{C_r}e^(iz)/(z-i)dz=0 rarr$
$rarr \int_{0}^{R}e^(ix)/(x-i)dx-1/e\int_{-1}^{-r}e^(-y)/ydy-1/e\int_{r}^{R-1}e^(-y)/ydy+\int_{C_R}e^(iz)/(z-i)dz+\int_{C_r}e^(iz)/(z-i)dz=0$
Passando al limite per $[r rarr 0^+]$ e per $[R rarr +oo]$:
$\int_{0}^{+oo}e^(ix)/(x-i)dx-1/eVP\int_{-1}^{+oo}e^(-y)/ydy-(\pii)/e=0 rarr$
$rarr \int_{0}^{+oo}e^(ix)/(x-i)dx=1/eVP\int_{-1}^{+oo}e^(-y)/ydy+(\pii)/e$
2. $f_2(z)=e^(iz)/(z+i)$
$\gamma_2$ percorsa in senso antiorario
$\int_{\gamma_2}e^(iz)/(z+i)dz=0 rarr$
$rarr \int_{0}^{R}e^(ix)/(x+i)dx-i\int_{0}^{R}e^(-t)/(it+i)dt+\int_{C_R}e^(iz)/(z+i)dz=0 rarr$
$rarr \int_{0}^{R}e^(ix)/(x+i)dx-e\int_{0}^{R}e^(-(t+1))/(t+1)dt+\int_{C_R}e^(iz)/(z+i)dz=0 rarr$
$rarr \int_{0}^{R}e^(ix)/(x+i)dx-e\int_{1}^{R+1}e^(-y)/ydy+\int_{C_R}e^(iz)/(z+i)dz=0$
Passando al limite per $[R rarr +oo]$:
$\int_{0}^{+oo}e^(ix)/(x+i)dx-e\int_{1}^{+oo}e^(-y)/ydy=0 rarr$
$rarr \int_{0}^{+oo}e^(ix)/(x+i)dx=e\int_{1}^{+oo}e^(-y)/ydy$
3. $f_3(z)=e^(-iz)/(z-i)$
$\gamma_3$ percorsa in senso orario
$\int_{\gamma_3}e^(-iz)/(z-i)dz=0 rarr$
$rarr \int_{0}^{R}e^(-ix)/(x-i)dx+i\int_{-R}^{0}e^t/(it-i)dt+\int_{C_R}e^(-iz)/(z-i)dz=0 rarr$
$rarr \int_{0}^{R}e^(-ix)/(x-i)dx+i\int_{0}^{R}e^(-t)/(-it-i)dt+\int_{C_R}e^(-iz)/(z-i)dz=0 rarr$
$rarr \int_{0}^{R}e^(-ix)/(x-i)dx-e\int_{0}^{R}e^(-(t+1))/(t+1)dt+\int_{C_R}e^(-iz)/(z-i)dz=0 rarr$
$rarr \int_{0}^{R}e^(-ix)/(x-i)dx-e\int_{1}^{R+1}e^(-y)/ydy+\int_{C_R}e^(-iz)/(z-i)dz=0$
Passando al limite per $[R rarr +oo]$:
$\int_{0}^{+oo}e^(-ix)/(x-i)dx-e\int_{1}^{+oo}e^(-y)/ydy=0 rarr$
$rarr \int_{0}^{+oo}e^(-ix)/(x-i)dx=e\int_{1}^{+oo}e^(-y)/ydy$
4. $f_4(z)=e^(-iz)/(z+i)$
$\gamma_4$ percorsa in senso orario
$\int_{\gamma_4}e^(-iz)/(z+i)dz=0 rarr$
$rarr \int_{0}^{R}e^(-ix)/(x+i)dx+i\int_{-R}^{-1-r}e^(t)/(it+i)dt+i\int_{-1+r}^{0}e^(t)/(it+i)dt+\int_{C_R}e^(-iz)/(z+i)dz+\int_{C_r}e^(-iz)/(z+i)dz=0 rarr$
$rarr \int_{0}^{R}e^(-ix)/(x+i)dx+i\int_{1+r}^{R}e^(-t)/(-it+i)dt+i\int_{0}^{1-r}e^(-t)/(-it+i)dt+\int_{C_R}e^(-iz)/(z+i)dz+\int_{C_r}e^(-iz)/(z+i)dz=0 rarr$
$rarr \int_{0}^{R}e^(-ix)/(x+i)dx-1/e\int_{1+r}^{R}e^(-(t-1))/(t-1)dt-1/e\int_{0}^{1-r}e^(-(t-1))/(t-1)dt+\int_{C_R}e^(-iz)/(z+i)dz+\int_{C_r}e^(-iz)/(z+i)dz=0 rarr$
$rarr \int_{0}^{R}e^(-ix)/(x+i)dx-1/e\int_{r}^{R-1}e^(-y)/ydy-1/e\int_{-1}^{-r}e^(-y)/ydy+\int_{C_R}e^(-iz)/(z+i)dz+\int_{C_r}e^(-iz)/(z+i)dz=0$
Passando al limite per $[r rarr 0^+]$ e per $[R rarr +oo]$:
$\int_{0}^{+oo}e^(-ix)/(x+i)dx-1/eVP\int_{-1}^{+oo}e^(-y)/ydy+(\pii)/e=0 rarr$
$rarr \int_{0}^{+oo}e^(-ix)/(x+i)dx=1/eVP\int_{-1}^{+oo}e^(-y)/ydy-(\pii)/e$
In definitiva:
$\int_{0}^{+oo}sinx/(x^2+1)dx=$
$=-1/4\int_{0}^{+oo}e^(ix)/(x-i)dx+1/4\int_{0}^{+oo}e^(ix)/(x+i)dx+1/4\int_{0}^{+oo}e^(-ix)/(x-i)dx-1/4\int_{0}^{+oo}e^(-ix)/(x+i)dx=$
$=-1/(4e)VP\int_{-1}^{+oo}e^(-y)/ydy-(\pii)/(4e)+e/4\int_{1}^{+oo}e^(-y)/ydy+e/4\int_{1}^{+oo}e^(-y)/ydy-1/(4e)VP\int_{-1}^{+oo}e^(-y)/ydy+(\pii)/(4e)=$
$=-1/(2e)VP\int_{-1}^{+oo}e^(-y)/ydy+e/2\int_{1}^{+oo}e^(-y)/ydy=$
$=(Ei(1)-e^2Ei(-1))/(2e)$