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Let $T$ be a positive real parameter. How does one evaluate the integral

$\displaystyle \int_0^{2 T^{1/4}} \sqrt{16T + x^4} dx$?

I checked "Handbook of Elliptic Integrals for Engineers and Scientists" by Byrd and Friedman, and it does not contain the evaluation of this type of integral, only the similar one

$\displaystyle \int_0^1 \frac{dx}{\sqrt{1 + x^4}}.$

Wolfram does give a closed form output for the indefinite integral, but it seems incomprehensible because the parameters are not real, and therefore it is not clear what it means for the definite integral.

syxiao
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1 Answers1

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Using the substitution $y=x/(2T^{1/4})$ we get $$\int_0^{2 T^{1/4}} \sqrt{16T + x^4} dx$$ $$=\int_0^1 2T^{1/4}\sqrt{16T+16y^4T}\,dy$$ $$=8T^{1/4}T^{1/2}\int_0^1 \sqrt{1+y^4}\,dy$$ the last integral can be expressed with the complete elliptic integral of the first kind $K$ $$=8T^{3/4}\frac{\sqrt{2}+K\left(\frac{\sqrt{2}}{2}\right)}{3}$$ $$=\frac{8}{3}T^{3/4}\left(\sqrt{2}+K\left(\frac{\sqrt{2}}{2}\right)\right)$$ With $K(\sqrt{2}/2) \approx 1.089429413224822$ you get $$\int_0^{2 T^{1/4}} \sqrt{16T + x^4} dx =8.71543530579858\cdot T^{3/4} $$

gammatester
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  • In fact $$\int_0^1 \sqrt {1 + {x^4}} dx$$ can be evaulated in terms of gamma function. See https://math.stackexchange.com/questions/2311583 – pisco Oct 27 '17 at 15:22