We have to find the Fourier transform of the following impulse response $$ h(t)=T[\delta (t)]=\frac{1}{2t_{d}}(\delta (t+t_{d})-\delta (t-t_{d})) $$
My own approach led to the following: $$ H(\omega )=\frac{1}{2t_{d}}\int_{-\infty }^{\infty }(\delta (t+t_{d})-\delta (t-t_{d}))e^{-i\omega t}dt $$ $$ H(\omega )=\frac{1}{2t_{d}}\int_{-\infty }^{\infty }(\delta (t+t_{d})e^{i\omega t_{d}}-\delta (t-t_{d})e^{-i\omega t_{d}})dt $$ $$ H(\omega )=\frac{1}{2t_{d}}(e^{i\omega t_{d}}-e^{-i\omega t_{d}}) $$ This however is wrong and the result should have been $$ H(\omega )=\frac{1}{2t_{d}}(e^{i\omega t_{d}}+e^{-i\omega t_{d}}) $$ I don't know what I did wrong so if anyone could show me that would help a lot.
I am getting the same result as you: $$ \int_{-\infty}^{\infty} \big(\delta (t+t_d) - \delta(t-t_d) \big)e^{-i\omega t} dt = \int_{-\infty}^{\infty}\delta (t+t_d) e^{-i\omega t} dt - \int_{-\infty}^{\infty} \delta(t-t_d) e^{-i\omega t} dt. $$ Then by doing changes of variables $u=t+t_d$ and $u=t-t_d$, you can express the integral as $$ \int_{-\infty}^{\infty}\delta (u) e^{-i\omega (u-t_d)} du - \int_{-\infty}^{\infty} \delta(u) e^{-i\omega (u+t_d)} du = e^{i\omega t_d} - e^{-i\omega t_d}.$$