I came across a textbook asking the following question:
If $h(t)$ is a real-valued function, show that the Fourier transform of $h(−t)$ is $H(f)^*$, where the asterisk ∗ denotes the complex conjugate.
Answer: $$\int_{-\infty}^{\infty}h(-t)e^{-j2\pi ft}dt = \int_{-\infty}^{\infty}h(\theta)e^{-j2\pi f(-\theta)}d{\theta} $$ $$=\int_{-\infty}^{\infty}h(\theta)e^{j2\pi f\theta}d{\theta}$$ $$=(\int_{-\infty}^{\infty}h(\theta)^*e^{j2\pi f\theta}d{\theta})^*$$ since $h(f)$ is real valued, $$=(\int_{-\infty}^{\infty}h(\theta)e^{j2\pi f\theta}d{\theta})^* $$ $$\int_{-\infty}^{\infty}h(-t)e^{-j2\pi ft}dt = H(f)^* $$
My question is: In the first step, $-t$ is substituted with $\theta$. However, shouldn't there be a "-" sign? as $dt=-d\theta$ during the substitution such that: $$\int_{-\infty}^{\infty}h(-t)e^{-j2\pi ft}dt = \int_{-\infty}^{\infty}-h(\theta)e^{-j2\pi f(-\theta)}d{\theta} $$
$$\int_{-\infty}^{\infty}h(-t)e^{-j2\pi ft}dt = \int_{\infty}^{-\infty} h(\theta) e^{-j2\pi f(-\theta)}(-d{\theta})\\ = \int_{-\infty}^{\infty} h(\theta) e^{-j2\pi f(-\theta)}d{\theta}\\ $$
The bounds flip and then the extra minus to flip them back.