I know the following integral representation of modified Bessel functions of first kind: $$I_q(\rho) = \frac{\left(\frac{\rho}{2}\right)^q}{\Gamma(q+1/2)\Gamma(1/2)} \int_{-1}^{1}e^{-\rho t}(1-t^2)^{q-1/2} dt $$ I wish to prove that $$I_{-q}(\rho)-I_q(\rho) = \frac{\Gamma(1/2-q)\sin(2q\pi)\left(\frac{\rho}{2}\right)^q}{\pi\Gamma(1/2)}\int_{1}^{\infty} e^{-\rho t}(t^2-1)^{q-1/2} dt$$
But I don't know which change of variable I must do to obtain this result, I don't understand how this integration limit change from $(-1\,\, \text{to}\,\, 1)$ to $(1\,\, \text{to}\,\, \infty$).
My final goal with this expressions is prove that $$K_q(\rho) = \frac{\Gamma(1/2)\left(\rho/2\right)^q}{\Gamma(q+1/2)}\int_{1}^{\infty} e^{-\rho t}(t^2-1)^{q-1/2} dt$$ where $K_q$ is the modified Bessel function of second kind which is defined for $q \not \in \mathbb{Z}$ as $$ \frac{\pi}{2}\frac{I_{-q}(\rho)-I_q(\rho)}{\sin(q \pi)}.$$
If anyone knows another way to prove the integral representation for $K_q$ above, it will very helpful, thank you so much for the attention.
The integral representation above can be found in the book A treatise on the theory of Bessel functions writed by Watson, page $172$
Update: The problem is solved if the following formula is proved for $\rho >0$ and $q \geq0$
$$I_{-q}(\rho) = \frac{(\rho/2)^q}{\sqrt{2\pi}\Gamma(q+1/2)}\left[ \frac{1}{2}\int_{-1}^{1} e^{\rho t}(1-t^2)^{q-1/2}dt+\sin(q \pi) \int_{1}^{\infty} e^{-\rho t}(t^2-1)^{q-1/2} dt\right]$$