I am trying to solve the following integral: \begin{equation}\label{1} \int \sin(\alpha t)\exp(-i\alpha t)\, \text{dt}. \end{equation} Using the exponential form of $\sin(\alpha t)$, the integral becomes \begin{equation}\label{pet} \int \frac{1}{2i}\big{(}\exp(i(\alpha-w)t)-\exp(-i(\alpha+w)t)\big{)}\,\text{dt} = \frac{1}{2i}\Big{(}\frac{1}{i(\alpha-w)}\exp(i(\alpha-w)t)+\frac{1}{i(\alpha+w)}\exp(-i(\alpha+w)t)\Big{)}\;\;(+\text{ constant}). \end{equation} When I input Eq. (2) into WolframAlpha, it returns True. If I rearrange the right-hand side of Eq. (2) by taking out $\frac{1}{i}$, I get \begin{equation} \frac{1}{2i}\Big{(}\frac{1}{i(\alpha-w)}\exp(i(\alpha-w)t)+\frac{1}{i(\alpha+w)}\exp(-i(\alpha+w)t)\Big{)} = -\frac{1}{2}\Big{(}\frac{1}{(\alpha-w)}\exp(i(\alpha-w)t)+\frac{1}{(\alpha+w)}\exp(-i(\alpha+w)t)\Big{)}. \end{equation} If I input Eq. (3) into WolframAlpha, it also returns True. However, when I enter the left-hand side of Eq. (2), i.e., the integral, equated with the new rearrangement from Eq. (3) as follows \begin{equation} \int \frac{1}{2i}\big{(}\exp(i(\alpha-w)t)-\exp(-i(\alpha+w)t)\big{)}\,\text{dt} = -\frac{1}{2}\Big{(}\frac{1}{(\alpha-w)}\exp(i(\alpha-w)t)+\frac{1}{(\alpha+w)}\exp(-i(\alpha+w)t)\Big{)}\;\;(+\text{ constant}), \end{equation} WolframAlpha returns False. I don't understand how this is possible. Surely, if $a = b \text{ and } b = c$, it must follow that $a = c$ by transitivity.
Here is the the code I typed directly into WolframAlpha:
- For Eq. (2): "is integral(sin(at)exp(-iwt)) = 1/(2i)(1/(i(a-w))exp(i(a-w)t)+1/(i(a+w))exp(-i(a+w)t))". This returns True.
- For Eq. (3): "is 1/(2i)(1/(i(a-w))exp(i(a-w)t)+1/(i(a+w))exp(-i(a+w)t)) = -1/2(1/(a-w)exp(i(a-w)t)+1/(a+w)exp(-i(a+w)t))". This also returns True.
- For Eq. (4): "is integral(sin(at)exp(-iwt)) = -1/2(1/(a-w)exp(i(a-w)t)+1/(a+w)exp(-i(a+w)t))". This returns False.
Thanks for your help.