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I've been trying to solve this integral and can't figure out where to start from: $$ \int\limits _0^1 r\sqrt{\dfrac{4r^2}{(r^2+\varepsilon^2)^2}+1} \textrm{ d}r $$ where $0<\varepsilon<1$.

Gepeto97
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  • Is $\varepsilon$ supposed to be small? If so you can expand the square root in a power series. $u=r^2$ looks like a bit of a simplification. – Ross Millikan Oct 13 '17 at 15:45
  • My bad, epsilon is bounded between 0 and 1 – Gepeto97 Oct 13 '17 at 15:47
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    I fed this to Alpha without success – Ross Millikan Oct 13 '17 at 15:55
  • The substitution $u=r^2+\epsilon^2$ immediately suggests itself, but is not enough on its own. Where did this integral come from? Do you have reasons to suspect it has a closed form? – user7530 Oct 13 '17 at 15:58
  • i would substitute $$t=\sqrt{\frac{4r^2}{(r^2+\epsilon^2)^2+1}}$$ – Dr. Sonnhard Graubner Oct 13 '17 at 15:58
  • According to Maple, it has a closed form, which is ... quite long. – nicomezi Oct 13 '17 at 16:01
  • maybe we can simplify the output – Dr. Sonnhard Graubner Oct 13 '17 at 16:03
  • Here is the output (a=$\epsilon$) : $-\ln(2)-\ln(a^2+1)+\ln(3+a^2+\sqrt{5+2a^2+a^4})-a\arctan\left(\dfrac{(\sqrt{5+2a^2+a^4}-a^2+1)a}{a^2\sqrt{5+2a^2+a^4}+a^2-1}\right)-(1/2)a^2+(1/2)\sqrt{5+2a^2+a^4}$ – nicomezi Oct 13 '17 at 16:06
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    @Dr.SonnhardGraubner -- your comment suggesting the substitution uses a radical expression that has a significant difference with the posted radical expression [Note: the +1 term placement]. Was this done on purpose? – K7PEH Oct 13 '17 at 16:29
  • I fed the integral into Mathematica and it produced an answer after about 15 to 20 seconds of processing (on MacBook Pro 2.2 GHz Intel Core i7). The answer is similar to what @nicomezi posted in his comment. I have not bothered to make sure they were identical or not, could be though. – K7PEH Oct 13 '17 at 16:32

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