Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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

$${I}\left({t}\right)=\int_{\mathrm{0}} ^{\:{t}} {e}^{{ix}\left(\pi+{i}\mathrm{ln}\:{x}\right)} {dx} \\ $$$$\mu=\pi+{i}\mathrm{ln}\:{x} \\ $$$$\therefore\:{x}={e}^{−{i}\left(\mu−\pi\right)} \:\:\:\Rightarrow\:\:\:{dx}={ie}^{−{i}\left(\mu−\pi\right)} {d}\mu \\ $$$$\: \\ $$$$\therefore{I}\left({t}\right)=\int_{−{i}\infty} ^{\:\pi+{i}\mathrm{ln}\:{t}} {e}^{{i}\mu{e}^{−{i}\left(\mu−\pi\right)} } {ie}^{−{i}\left(\mu−\pi\right)} {d}\mu \\ $$$${I}\left({t}\right)=\int_{−{i}\infty} ^{\:\pi+{i}\mathrm{ln}\:{t}} {ie}^{{i}\mu{e}^{−{i}\left(\mu−\pi\right)} −{i}\left(\mu−\pi\right)} {d}\mu \\ $$$${I}\left({t}\right)=\int_{−{i}\infty} ^{\:\pi+{i}\mathrm{ln}\:{t}} {ie}^{{i}\left(\mu{e}^{−{i}\left(\mu−\pi\right)} −\left(\mu−\pi\right)\right)} {d}\mu \\ $$$${I}\left({t}\right)=\int_{−{i}\infty} ^{\:\pi+{i}\mathrm{ln}\:{t}} {i}\left(\mathrm{cos}\left(\mu{e}^{−{i}\left(\mu−\pi\right)} −\left(\mu−\pi\right)\right)+{i}\mathrm{sin}\left(\mu{e}^{−{i}\left(\mu−\pi\right)} −\left(\mu−\pi\right)\right)\right){d}\mu \\ $$$${I}\left({t}\right)=\int_{−{i}\infty} ^{\:\pi+{i}\mathrm{ln}\:{t}} {i}\mathrm{cos}\left(\mu{e}^{−{i}\left(\mu−\pi\right)} −\mu+\pi\right){d}\mu−\int_{−{i}\infty} ^{\:\pi+{i}\mathrm{ln}\:{t}} \mathrm{sin}\left(\mu{e}^{−{i}\left(\mu−\pi\right)} −\mu+\pi\right){d}\mu \\ $$$${e}^{−{i}\left(\mu−\pi\right)} =−{e}^{−{i}\mu} \\ $$$${I}\left({t}\right)=\int_{−{i}\infty} ^{\:\pi+{i}\mathrm{ln}\:{t}} {i}\mathrm{cos}\left(−\mu{e}^{−{i}\mu} −\mu+\pi\right){d}\mu−\int_{−{i}\infty} ^{\:\pi+{i}\mathrm{ln}\:{t}} \mathrm{sin}\left(−\mu{e}^{−{i}\mu} −\mu+\pi\right){d}\mu \\ $$$$\mathrm{Please}\:\mathrm{continue} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com