Im supposed to show that $C_n = \frac{1(1-e^{-\pi}(-1)^n)}{\pi(1+n^2)}$
I'm computing the Fourier coefficients of the $2\pi$-periodic function $f(x) = \exp(-|x|)$ on $[-\pi,\pi]$. I need to show that: $$\hat{f}(n) = \frac{1-e^{-\pi}(-1)^n}{\pi(1+n^2)}$$
So far, this is what I have done:
$$\hat{f}(n) = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) e^{-inx} \ dx = \frac{1}{2\pi} \int_{-\pi}^{0} e^x e^{-inx} \ dx + \frac{1}{2\pi} \int_{0}^{\pi} e^{-x} e^{-inx} \ dx$$
Then, we have that: $$\int_{-\pi}^{0} e^x e^{-inx} \ dx = \int_{-\pi}^{0} e^{x-inx} \ dx = \int_{-\pi}^{0} e^{x(1-in)} \ dx \frac{1}{1-in}(1-e^{-\pi(1-in)})$$
$$\int_{0}^{\pi} e^{-x} e^{-inx} \ dx = \int_{0}^{\pi} e^{-x(1+in)} \ dx = -\frac{1}{1+in} (e^{-\pi(1+in)}-1) = \frac{1}{1+in} (1-e^{-\pi(1+in)})$$
So, then: $$\hat{f}(n) = \frac{1}{2\pi} \frac{1}{1-in} (1-e^{-\pi(1-in)}) + \frac{1}{2\pi} \frac{1}{1+in} (1-e^{-\pi(1+in)})$$
I have got so far, but I do not how to move forward to remove the $i$s.
First, let's recognize that: $$\exp(i\pi n) = (\exp(i\pi))^n = (-1)^n$$
Right, so you have that: $$2\pi c_n(1+n^2) = (1+in)(1-\exp(-\pi)(-1)^n) + (1-in)(1-\exp(-\pi)(-1)^n) = (1+in+1-in)(1-\exp(-\pi)(-1)^n)$$ I just multiplied throughout by $1+n^2$ and used the fact that $1+n^2 = (1-in)(1+in)$. Also, I've used the fact that: $$\exp(-\pi(1-in)) = \exp(-\pi)\exp(i\pi n) = \exp(-\pi) (-1)^n$$ and I've done the same thing for the other term (this means that you should do the calculation yourself and verify that it's correct). So, it follows that: $$2\pi c_n(1+n^2) = 2(1-\exp(-\pi)(-1)^n)$$
But this implies that: $$c_n = \frac{1-\exp(-\pi)(-1)^n}{\pi(1+n^2)}$$ and you're done. For future reference, you should try to type out your work here in latex so that your questions don't get downvoted. It's just fine in this case (at least, for me) because your handwriting is gorgeous :)