I think about on this and found a result, [tex]e^{i\pi k}=e^{-i\pi k}=-1\Rightarrow i\pi k=ln(-1)[/tex](adsbygoogle = window.adsbygoogle || []).push({});

a>1 and k=1,2,3...

[tex]y=(-a)^{i\alpha }\Rightarrow lny=i\alpha ln(-a)=i\alpha (ln(-1)+lna)=i\alpha (i\pi k+lna)\Rightarrow y=e^{-\alpha \pi k}e^{i\alpha \ln{a}}[/tex]

Then

[tex]\alpha \ln{a}=-\pi \Rightarrow \alpha =\frac{-\pi }{\ln{a}}\Rightarrow e^{-i\pi }e^{\frac{{\pi }^2 k}{\ln{a}}}=-e^{\frac{{\pi }^2 k}{\ln{a}}} [/tex]

If I choose k as infinity, I get a exponential expression as negative infinity.

Is it right? Please explain to me. Thanks

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# Can a^x be negative infinity

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