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Starting from the equation, where j represents an imaginary number $j = \sqrt{-1}$:

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To reach the answer:

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I wrote the first equation as $d[n] = \frac{1}{2*pi} \int j \frac{j \Omega}{\pi}e^{(n+3/2)\Omega}$. Then chose $u=\frac{1}{2*pi}$ and $dv = e^{(n+3/2)\Omega}$

What I get is $d[n] = \frac{1}{2\pi}[\frac{j\Omega}{pi}\frac{1}{n+3/2}e^{(n+3/2)\Omega} - \frac{j}{\pi}\frac{1}{(n+3/2)^2}e^{(n+3/2)\Omega}$ I have to apply eulers formula to get the sin and cosine, but I dont think I got the correct answer to do that.

Thank you.

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Your integration by parts is a bit off. You correctly rewrote $d[n]$; after factoring out constants, we get that

$$d[n] = \frac{1}{2\pi} \frac{j}{\pi} \int j \cdot e^{\frac{1}{2}j (2n-3) \Omega} d\Omega$$

Now, setting $u=x$ and $dv=e^{\frac{1}{2}j (2n-3) \Omega} d\Omega$, you can integrate by parts to get

$$d[n] = \left[2 \Omega e^{\frac{1}{2}j(2n-3)\Omega} - 2\int e^{\frac{1}{2}j(2n-3)\Omega}d\Omega\right]$$

The inner integral evaluates to

$$\frac{-2je^{\frac{1}{2}j(2n-3)\Omega}}{2n-3}$$

(you can obtain this by $u$-substitution of $\frac{j}{2}(2n-3)\Omega$). When we plug this back in and simplify, we get that

$$d[n] = \left(\frac{1}{\pi} \cdot \left(j e^{\frac{1}{2}j(2n-3)}\right) \cdot \left(\frac{4}{(2n-3)^2} - \frac{2j\Omega}{2n-3}\right)\right)\Big|_a^b$$

which, I believe, should be exactly equal to the answer above after applying Euler's formula, as you said, to $e^{\frac{1}{2}j(2n-3)}$.

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