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Question Number 6468 by Temp last updated on 28/Jun/16
I(t)=∫_0 ^( t) e^(iπx) x^(−x) dx, t∈Z
$${I}\left({t}\right)=\int_{\mathrm{0}} ^{\:{t}} {e}^{{i}\pi{x}} {x}^{−{x}} {dx},\:\:{t}\in\mathbb{Z} \\ $$
Commented by Temp last updated on 28/Jun/16
I(t)=∫_0 ^( t) e^(iπx) e^(−xln x) dx I(t)=∫_0 ^( t) e^(x(iπ−ln x)) dx I(t)=∫_0 ^( t) e^(xi(π+iln x)) dx I(t)=∫_0 ^( t) (cos x + isin x)^((π+iln x)) dx
$${I}\left({t}\right)=\int_{\mathrm{0}} ^{\:{t}} {e}^{{i}\pi{x}} {e}^{−{x}\mathrm{ln}\:{x}} {dx} \\ $$$${I}\left({t}\right)=\int_{\mathrm{0}} ^{\:{t}} {e}^{{x}\left({i}\pi−\mathrm{ln}\:{x}\right)} {dx} \\ $$$${I}\left({t}\right)=\int_{\mathrm{0}} ^{\:{t}} {e}^{{xi}\left(\pi+{i}\mathrm{ln}\:{x}\right)} {dx} \\ $$$${I}\left({t}\right)=\int_{\mathrm{0}} ^{\:{t}} \left(\mathrm{cos}\:{x}\:+\:{i}\mathrm{sin}\:{x}\right)^{\left(\pi+{i}\mathrm{ln}\:{x}\right)} {dx} \\ $$
Commented by Temp last updated on 28/Jun/16
I(t)=∫_0 ^( t) e^(ix(π+iln x)) dx μ=π+iln x ∴ x=e^(−i(μ−π)) ⇒ dx=ie^(−i(μ−π)) dμ ∴I(t)=∫_(−i∞) ^( π+iln t) e^(iμe^(−i(μ−π)) ) ie^(−i(μ−π)) dμ I(t)=∫_(−i∞) ^( π+iln t) ie^(iμe^(−i(μ−π)) −i(μ−π)) dμ I(t)=∫_(−i∞) ^( π+iln t) ie^(i(μe^(−i(μ−π)) −(μ−π))) dμ I(t)=∫_(−i∞) ^( π+iln t) i(cos(μe^(−i(μ−π)) −(μ−π))+isin(μe^(−i(μ−π)) −(μ−π)))dμ I(t)=∫_(−i∞) ^( π+iln t) icos(μe^(−i(μ−π)) −μ+π)dμ−∫_(−i∞) ^( π+iln t) sin(μe^(−i(μ−π)) −μ+π)dμ e^(−i(μ−π)) =−e^(−iμ) I(t)=∫_(−i∞) ^( π+iln t) icos(−μe^(−iμ) −μ+π)dμ−∫_(−i∞) ^( π+iln t) sin(−μe^(−iμ) −μ+π)dμ Please continue
$${I}\left({t}\right)=\int_{\mathrm{0}} ^{\:{t}} {e}^{{ix}\left(\pi+{i}\mathrm{ln}\:{x}\right)} {dx} \\ $$$$\mu=\pi+{i}\mathrm{ln}\:{x} \\ $$$$\therefore\:{x}={e}^{−{i}\left(\mu−\pi\right)} \:\:\:\Rightarrow\:\:\:{dx}={ie}^{−{i}\left(\mu−\pi\right)} {d}\mu \\ $$$$\: \\ $$$$\therefore{I}\left({t}\right)=\int_{−{i}\infty} ^{\:\pi+{i}\mathrm{ln}\:{t}} {e}^{{i}\mu{e}^{−{i}\left(\mu−\pi\right)} } {ie}^{−{i}\left(\mu−\pi\right)} {d}\mu \\ $$$${I}\left({t}\right)=\int_{−{i}\infty} ^{\:\pi+{i}\mathrm{ln}\:{t}} {ie}^{{i}\mu{e}^{−{i}\left(\mu−\pi\right)} −{i}\left(\mu−\pi\right)} {d}\mu \\ $$$${I}\left({t}\right)=\int_{−{i}\infty} ^{\:\pi+{i}\mathrm{ln}\:{t}} {ie}^{{i}\left(\mu{e}^{−{i}\left(\mu−\pi\right)} −\left(\mu−\pi\right)\right)} {d}\mu \\ $$$${I}\left({t}\right)=\int_{−{i}\infty} ^{\:\pi+{i}\mathrm{ln}\:{t}} {i}\left(\mathrm{cos}\left(\mu{e}^{−{i}\left(\mu−\pi\right)} −\left(\mu−\pi\right)\right)+{i}\mathrm{sin}\left(\mu{e}^{−{i}\left(\mu−\pi\right)} −\left(\mu−\pi\right)\right)\right){d}\mu \\ $$$${I}\left({t}\right)=\int_{−{i}\infty} ^{\:\pi+{i}\mathrm{ln}\:{t}} {i}\mathrm{cos}\left(\mu{e}^{−{i}\left(\mu−\pi\right)} −\mu+\pi\right){d}\mu−\int_{−{i}\infty} ^{\:\pi+{i}\mathrm{ln}\:{t}} \mathrm{sin}\left(\mu{e}^{−{i}\left(\mu−\pi\right)} −\mu+\pi\right){d}\mu \\ $$$${e}^{−{i}\left(\mu−\pi\right)} =−{e}^{−{i}\mu} \\ $$$${I}\left({t}\right)=\int_{−{i}\infty} ^{\:\pi+{i}\mathrm{ln}\:{t}} {i}\mathrm{cos}\left(−\mu{e}^{−{i}\mu} −\mu+\pi\right){d}\mu−\int_{−{i}\infty} ^{\:\pi+{i}\mathrm{ln}\:{t}} \mathrm{sin}\left(−\mu{e}^{−{i}\mu} −\mu+\pi\right){d}\mu \\ $$$$\mathrm{Please}\:\mathrm{continue} \\ $$

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