# A Question on the Application of Integration by Parts

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

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.

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 $$e^{\frac{1}{2}j(2n-3)}$$.