Is it possible to write $\int_{-\infty}^\infty e^{-iqu}\left( \frac{A}{\sqrt{1-iau}}+\frac{B}{\sqrt{1-ibu}}\right)^Ndu$ in this form $\int\frac{e^{-iu}}{(\sqrt{1-iau})^n(\sqrt{1-ibu})^m}du$
Writing an Integral in Different Form?
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$\begingroup$
integration
transformation
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0Binomium formula? – 2011-01-18
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0tat makes it more complicated...?? – 2011-01-18
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0But that's precisely what you want. – 2011-01-18