Integral Problem!!

For $\displaystyle\int x^{i} \:\ dx :$

$\displaystyle\int x^{i} \:\ dx = \displaystyle\int e^{i\ln(x)} \:\ dx$

Let $u = e^{i\ln(x)} \implies u' = \dfrac{i}{x}e^{i\ln(x)}$ and $v' = 1 \implies v = x \implies$

$\displaystyle\int e^{i\ln(x)} \: dx = x e^{i\ln(x)} - i\displaystyle\int e^{i\ln(x)} \:\ dx \implies$

$(1 + i)\displaystyle\int e^{i\ln(x)} \:\ dx = x e^{i\ln(x)} \implies$

$\displaystyle\int e^{i\ln(x)} \: dx = \dfrac{1}{1 + i}x e^{i\ln(x)} = \dfrac{1}{2}(1 - i) x e^{i\ln(x)}$

$\implies$

$\displaystyle\int_{e^{-\pi}}^{e^{\pi}} x^{i} \:\ dx =$ $\dfrac{1}{2}(1 - i)(e^{\pi} * e^{i\pi} - e^{-\pi} * e^{-i\pi}) = \dfrac{1}{2}(1 - i)(-e^{\pi} + e^{-\pi})$

$= \dfrac{1}{2}(1 - i)(e^{-\pi} - e^{\pi})$

$\implies (\displaystyle\int_{e^{-\pi}}^{e^{\pi}} x^{i} \:\ dx) * (-1 - i) =$ $-\dfrac{1}{2}(1 - i^2)(e^{-\pi} - e^{\pi}) = e^{\pi} - e^{-\pi} \approx \boxed{23.0974787}$ .