I w'd surprised if this integral $\int\tan ( e^{-x²})\ d x$ has a closed form since :$\int_{-\infty}^{+\infty }\tan ( e^{-x²})\ d x$ is assumed as a constant by wolfram alpha and The inverse calculator didn't give me any thing such that it's value is $2.27591....$
, then any way to show that if it has a closed form , or it is just a constant ?
Sunday, 23 November 2014
integration - Closed form of :$inttan ( e^{-x²}) d x$ Over reals
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