1

How can I evaluate this integral? $$\int_{0}^{1} \frac{dx}{\sqrt{{1+x^4} }}$$ I tried using the substitution $x=\mathrm{e}^{-u}$ but I got nowhere.

user281270
  • 83
  • @Semiclassical: not really, because the easiest way, as shown in math110's answer, to handle that question is to not evaluate this integral. However, looking back at that question, I see that my answer here is very close to my answer there. – robjohn Mar 03 '16 at 22:53

1 Answers1

2

$$ \begin{align} \int_0^1\frac{\mathrm{d}x}{\sqrt{1+x^4}} &=\int_1^\infty\frac{\mathrm{d}x}{\sqrt{1+x^4}}\tag{1}\\ &=\frac12\int_0^\infty\frac{\mathrm{d}x}{\sqrt{1+x^4}}\tag{2}\\ &=\frac12\int_0^\infty\frac14x^{-3/4}(1+x)^{-1/2}\,\mathrm{d}x\tag{3}\\[3pt] &=\frac18\mathrm{B}\left(\frac14,\frac14\right)\tag{4}\\ &=\frac{\Gamma\left(\frac14\right)^2}{8\sqrt\pi}\tag{5} \end{align} $$ Explanation:
$(1)$: substitute $x\mapsto\frac1x$
$(2)$: average left and right sides of $(1)$
$(3)$: substitute $x\mapsto x^{1/4}$
$(4)$: Beta integral
$(5)$: translate to Gamma functions

robjohn
  • 345,667