Please, could anyone help me to prove that this equality holds? I compute some integrals, using residuals and I know the result, but I am not able to recompose my result to the one which is on the right side.
$$ i\pi \frac {1 - i - i \sqrt{2}}{8e^{\frac{1}{8}i\pi}} = \frac{\pi}{8 \sin \frac{\pi}{8}} $$
Thank you for your help.
Hint
$$1-\cos(\frac{\pi}{4})=2\sin^2(\frac{\pi}{8})$$
$$1+\cos(\frac{\pi}{4})=2\cos^2(\frac{\pi}{8})$$
$$cos(\frac{\pi}{8})=(1+\sqrt{2})\sin(\frac{\pi}{8})$$
Set $\pi/8=\alpha$ for convenience. What you want to see is that $$ (1+\sqrt{2}+i)\sin\alpha=\cos\alpha+i\sin\alpha $$ that is, $$ \cos\alpha=(1+\sqrt{2})\sin\alpha $$ which is the same as $$ \cos^2\alpha=(3+2\sqrt{2})\sin^2\alpha $$ or $$ \sin^2\alpha=\frac{1}{4+2\sqrt{2}} $$ that becomes $$ \frac{1-\cos2\alpha}{2}=\frac{1}{4+2\sqrt{2}} $$ or $$ \cos2\alpha=1-\frac{1}{2+\sqrt{2}}= \frac{1+\sqrt{2}}{2+\sqrt{2}}=\frac{1}{\sqrt{2}} $$ which is true. Since all steps are reversible, you get your identity (the step where I squared both sides is reversible because both terms are positive).
EDIT: Ah, I see your issue has already been resolved, but I'll just be leaving this here since I already spent the time on it.
After working through this, I'm sure there's another way that's shorter, especially since the last few steps are extremely counterintuitive. But here's one way.
\begin{align*} i \pi \frac{1-i-i\sqrt2}{8e^{i\pi/8}} &= \frac\pi8 ie^{-i\pi/8}[1 - i(1+\sqrt2)]\\[0.3cm] &= \frac\pi8 i\left(\cos \frac\pi8 - i\sin\frac\pi8\right)[1 - i(1+\sqrt2)]\\[0.3cm] &= \frac\pi8 i \left(\cos\frac\pi8 - i(1 + \sqrt2)\cos\frac\pi8 - i\sin\frac\pi8 + i^2 (1+\sqrt2)\sin\frac\pi8 \right)\\[0.3cm] &= \frac\pi8 i \left(\cos\frac\pi8 - i(1 + \sqrt2)\cos\frac\pi8 - i\sin\frac\pi8 - (1+\sqrt2)\sin\frac\pi8 \right)\\[0.3cm] &= \frac\pi8 \left(i\cos\frac\pi8 - i^2(1 + \sqrt2)\cos\frac\pi8 - i^2\sin\frac\pi8 - i(1+\sqrt2)\sin\frac\pi8 \right)\\[0.3cm] &= \frac\pi8 \left(i\cos\frac\pi8 + (1 + \sqrt2)\cos\frac\pi8 + \sin\frac\pi8 - i(1+\sqrt2)\sin\frac\pi8 \right)\\[0.3cm] &= \frac\pi8 \left((1+\sqrt2)\cos\frac\pi8 + \sin\frac\pi8 + i\left[\cos\frac\pi8 - (1+\sqrt2)\sin\frac\pi8\right]\right) \end{align*}
FUN FACT: $\displaystyle \cos\frac\pi8 = (1+\sqrt2)\sin\frac\pi8$
PROOF OF FUN FACT: \begin{align*} \frac{\cos\frac\pi8}{\sin\frac\pi8} &= \cot\frac\pi8\\[0.3cm] &= \cot\left(\frac12 \cdot \frac\pi4\right)\\[0.3cm] &= \cot \frac\pi4 + \csc\frac\pi4\\[0.3cm] &= 1 + \sqrt2 \end{align*} Therefore $\displaystyle \cos\frac\pi8 = (1 + \sqrt2)\sin\frac\pi8$. Substitute this into the last line of our chain of equations above and we get \begin{align*} i \pi \frac{1-i-i\sqrt2}{8e^{i\pi/8}} &= \cdots\\ &= \frac\pi8 \left((1+\sqrt2)\cos\frac\pi8 + \sin\frac\pi8 + i\left[\cos\frac\pi8 - (1+\sqrt2)\sin\frac\pi8\right]\right)\\[0.3cm] &= \frac\pi8 \left( (1+\sqrt2)(1+\sqrt2)\sin\frac\pi8 + \sin\frac\pi8 + i\left[(1+\sqrt2)\sin\frac\pi8 - (1+\sqrt2)\sin\frac\pi8)\right]\right)\\[0.3cm] &= \frac\pi8 \left( (3+2\sqrt2)\sin\frac\pi8 + \sin\frac\pi8 + i \cdot 0\right)\\[0.3cm] &= \frac\pi8 (4+2\sqrt2)\sin\frac\pi8\\[0.3cm] &= \frac\pi4 (2+\sqrt2)\sin\frac\pi8 \end{align*}
If it were me I would've stopped here (at the latest) but since your question is how is your answer equivalent to $\dfrac\pi{8\sin\frac\pi8}$ then we'll continue: \begin{align*} i \pi \frac{1-i-i\sqrt2}{8e^{i\pi/8}} &= \cdots\\ &= \frac\pi4 (2+\sqrt2)\sin\frac\pi8 \cdot \color{red}{\frac{\sin\frac\pi8}{\sin\frac\pi8}}\\[0.3cm] &= \frac\pi4 (2+\sqrt2)\sin^2 \frac\pi8 \cdot \frac1{\sin\frac\pi8}\\[0.3cm] &= \frac\pi4 (2+\sqrt2)\left[\sin^2 \left(\frac12 \cdot \frac\pi4\right)\right] \cdot \frac1{\sin\frac\pi8}\\[0.3cm] &= \frac\pi4 (2+\sqrt2)\left[\frac{1-\cos\frac\pi4}2\right] \cdot \frac1{\sin\frac\pi8}\\[0.3cm] &= \frac\pi8 (2+\sqrt2)\left(1-\frac{\sqrt2}2\right) \cdot \frac1{\sin\frac\pi8}\\[0.3cm] &= \frac\pi8 \left(2 - \sqrt2 + \sqrt2 - \frac22\right) \cdot \frac1{\sin\frac\pi8}\\[0.3cm] &= \frac\pi{8\sin\frac\pi8} \end{align*}
One approach uses the fact that $\sin(\theta)=\dfrac{e^{i\theta}-e^{-i\theta}}{2i}$. After some simple manipulations, we get to
$$(i+1+\sqrt{2})\left(\frac{e^{i\pi/8}-e^{-i\pi/8}}{2i}\right)=e^{i\pi/8}$$
$$(1+\sqrt{2}+i)e^{i\pi/8}+(1+\sqrt{2}+i)\frac{1}{e^{i\pi/8}}=2ie^{i\pi/8}$$
$$(1+\sqrt{2}+i)\frac{1}{e^{i\pi/8}}=(-1-\sqrt{2}+i)e^{i\pi/8}$$
$$1+\sqrt{2}+i=(-1-\sqrt{2}+i)e^{i\pi/4}$$
Now use that $e^{i\pi/4}=\cos(\tfrac{\pi}{4})+i\sin(\tfrac{\pi}{4})=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i$.