MHB Double Integral A₁, -1 - Does it Check Out?

[Dustinsfl]

\begin{alignat*}{3}
A_{1,-1} & = & \frac{50\sqrt{3}}{\sqrt{2\pi}}\int_{-\pi/4}^{\pi/4}\int_0^{\pi}e^{-i\varphi}\sin\theta d\theta d\varphi\\
& = & \frac{100\sqrt{3}}{\sqrt{2\pi}}\int_{-\pi/4}^{\pi/4}e^{-i\varphi}d\varphi\\
& = & \frac{100\sqrt{3}}{\sqrt{\pi}}\\
\end{alignat*}

Is this correct?

 

[Dustinsfl]

\begin{alignat*}{3}
A_{1,-1} & = & \frac{50\sqrt{3}}{\sqrt{2\pi}}\int_{-\pi/4}^{\pi/4}\int_0^{\pi}e^{-i\varphi}\sin\theta d\theta d\varphi\\
& = & \frac{100\sqrt{3}}{\sqrt{2\pi}}\int_{-\pi/4}^{\pi/4}e^{-i\varphi}d\varphi\\
& = & \frac{100\sqrt{3}}{\sqrt{\pi}}\\
\end{alignat*}

Is this correct?

No need to answer this question. I was asking this in reference to something I was doing in Spherical Harmonics. I found the error of my way.