What would be the evaluation of the definite integral $\int\limits_1^\infty x^{y^m}dy$ where $x$ with $0 < x < 1 $ is real ; $m > 1$ is any integer. At least an approximation will suffice if the error of the approximation is determinable.
Evaluate integral $\int\limits_1^\infty x^{y^m}dy$
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real-analysis
definite-integrals
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1See [Generalized error functions](http://en.wikipedia.org/wiki/Error_function#Generalized_error_functions) (non-standard) – 2012-11-02
2 Answers
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Using the incomplete gamma function, this is $ \frac{\Gamma(1/m,-\log x)}{m\cdot(-\log x)^{1/m}}. $ Hint: Use the change of variable $t=-\log x\cdot y^m$.
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Since $x \in (0,1)$ this is the same as $1/a$ where $a\in(0,\infty)$ then by some numerical calculations I obtained that $ \int_0^\infty \left(\frac{1}{a}\right)^{y^r}\mathrm{d}y=-\frac{1}{r} \cdot \frac{(-1)^{1 - 2/r} \Gamma(1/r)}{\sqrt[r]{\log(a)}} $ So now you only have to subtract the value of $ \int_0^1 \left( \frac{1}{a}\right)^{y^r} \mathrm{d}y$ to get your desired integral. The latter integral can be approximated in numerous ways. Although I think a taylor expansion or a sum would suffice.
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0dunno, why you askin me bout that? – 2012-11-02