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Question Number 6468 by Temp last updated on 28/Jun/16

$$I\left(t\right)={\int }_0^t{e}^{i\pi x}{x}^{-x}dx,t\in \mathbb{Z}$$

Commented by Temp last updated on 28/Jun/16

$$I\left(t\right)={\int }_0^t{e}^{i\pi x}{e}^{-x\ln x}dx$$$$I\left(t\right)={\int }_0^t{e}^{x(i\pi -\ln x)}dx$$$$I\left(t\right)={\int }_0^t{e}^{xi(\pi +i\ln x)}dx$$$$I\left(t\right)={\int }_0^t{(\cos x+i\sin x)}^{(\pi +i\ln x)}dx$$

Commented by Temp last updated on 28/Jun/16

$$I\left(t\right)={\int }_0^t{e}^{ix(\pi +i\ln x)}dx$$$$\mu =\pi +i\ln x$$$$\therefore x=e^{-i(\mu -\pi )}\Rightarrow dx={ie}^{-i(\mu -\pi )}d\mu$$$$$$$$\therefore I\left(t\right)={\int }_{-i\mathrm{\infty }}^{\pi +i\ln t}{e}^{i\mu e^{-i(\mu -\pi )}}{ie}^{-i(\mu -\pi )}d\mu$$$$I\left(t\right)={\int }_{-i\mathrm{\infty }}^{\pi +i\ln t}{ie}^{i\mu e^{-i(\mu -\pi )}-i(\mu -\pi )}d\mu$$$$I\left(t\right)={\int }_{-i\mathrm{\infty }}^{\pi +i\ln t}{ie}^{i(\mu e^{-i(\mu -\pi )}-(\mu -\pi ))}d\mu$$$$I\left(t\right)={\int }_{-i\mathrm{\infty }}^{\pi +i\ln t}i(\cos (\mu e^{-i(\mu -\pi )}-(\mu -\pi ))+i\sin (\mu e^{-i(\mu -\pi )}-(\mu -\pi )))d\mu$$$$I\left(t\right)={\int }_{-i\mathrm{\infty }}^{\pi +i\ln t}i\cos (\mu e^{-i(\mu -\pi )}-\mu +\pi )d\mu -{\int }_{-i\mathrm{\infty }}^{\pi +i\ln t}\sin (\mu e^{-i(\mu -\pi )}-\mu +\pi )d\mu$$$$e^{-i(\mu -\pi )}=-e^{-i\mu }$$$$I\left(t\right)={\int }_{-i\mathrm{\infty }}^{\pi +i\ln t}i\cos (-\mu e^{-i\mu }-\mu +\pi )d\mu -{\int }_{-i\mathrm{\infty }}^{\pi +i\ln t}\sin (-\mu e^{-i\mu }-\mu +\pi )d\mu$$$$\mathrm{Please}\mathrm{continue}$$

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