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

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SUMMARY

The discussion centers on the evaluation of the double integral A₁, -1, specifically the expression A₁, -1 = (50√3/√(2π))∫(-π/4)^(π/4)∫0^(π)e^(-iφ)sin(θ) dθ dφ. The integral simplifies to A₁, -1 = (100√3/√(2π))∫(-π/4)^(π/4)e^(-iφ)dφ, ultimately resulting in A₁, -1 = (100√3/√π). The correctness of this evaluation was confirmed in the context of Spherical Harmonics, indicating a resolution of prior confusion regarding the integral's computation.

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\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?
 
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dwsmith said:
\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.
 

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