Response of an LTI system by convolution
In the convolution integral we're used to seeing \$h(t - \tau)\$ when talking about impulse response \$h(t)\$, however:

$$\int_{-\infty}^\infty{f(\tau)}g(t - \tau)d\tau = \int_{-\infty}^\infty{f(t - \tau)}g(\tau)d\tau $$

In this case the left side is used with \$ f(\tau) = h(\tau)\$ and \$ g(t - \tau) = e^{j\omega_k(t - \tau)}\$ so you have:

$$\int_{-\infty}^\infty{f(\tau)}g(t - \tau)d\tau = \int_{-\infty}^\infty{h(\tau)}e^{j\omega_k(t - \tau)}d\tau = e^{j\omega_kt}\int_{-\infty}^\infty{h(\tau)}e^{-j\omega_k\tau}d\tau$$