by Chang Henry
Last Updated October 20, 2019 05:20 AM - source

I have got an definite integral and I think it is impossible to the exact value since it is not elementary. The question says that ε positive but small, so I believe it must be an approximation. Then the problem lays on approximation. I have to integrate root of

$$\sqrt{a^2-x^2+(2ε/μ^2)(a^4-x^4)}$$

from upper limit $a$ to low limit 0.(Forgive me for being unable to type the proper symbols) If the ε is 0, then we get an $\pi/2$ which is clear enough. But then the question asks to find the value of integral reduced by when ε is small and greater than 0. I don't know how to use the condition ε is small. Certainly it can't simply be zero, so what should I do?

Approximate the integrand as follows, treating $\epsilon$ as a small perterbation variable,

$$\sqrt{a^2-x^2+(2ε/μ^2)(a^4-x^4)}$$ $$=\sqrt{(a^2-x^2)[1+(2ε/μ^2)(a^2+x^2)]}$$ $$\approx \sqrt{a^2-x^2}\left[1+(ε/μ^2)(a^2+x^2)\right]$$

Then, the original integral is approximated as,

$$I= \int_0^a \sqrt{a^2-x^2}dx+(ε/μ^2)\int_0^a \sqrt{a^2-x^2}(a^2+x^2)dx$$

Now, the two integrals above can be carried out separately. Note that the first integral is the one as if with $\epsilon$ set to zero, and the second integral is the correction term due to small but non-vanishing $\epsilon$. Its value is

$$\int_0^a \sqrt{a^2-x^2}(a^2+x^2)dx = \frac{5\pi}{16}a^4$$

October 19, 2019 15:41 PM

Using @Quanto's answer, you could push the approximation almost as far as you want.

For conveniency, let $c=\frac {2\epsilon} {\mu^2}$ and,as Quanto wrote $$\sqrt{a^2-x^2+c(a^4-x^4)}=\sqrt{(a^2-x^2)[1+c(a^2+x^2)]}=\sqrt{a^2-x^2}\sqrt{1+c(a^2+x^2)}$$ Now, using the binomial expansion $$\sqrt{1+c(a^2+x^2)}=\sum_{n=0}^\infty \binom{\frac{1}{2}}{n}c^n \left(a^2+x^2\right)^n$$ $$\int_0^a\sqrt{a^2-x^2+c(a^4-x^4)}\,dx=\sum_{n=0}^\infty \binom{\frac{1}{2}}{n}c^n \int_0^a\sqrt{a^2-x^2}\left(a^2+x^2\right)^n\,dx=\sum_{n=0}^\infty \binom{\frac{1}{2}}{n}c^n I_n$$ $$I_n=\int_0^a\sqrt{a^2-x^2}\left(a^2+x^2\right)^n\,dx=a^{2 n+2}\int_0^1\sqrt{1-t^2} \left(1+t^2\right)^n\,dt=a^{2 n+2}J_n$$ and, for the first values of $n$, the $J_n$ correspond to the sequence

$$\left\{\frac{\pi }{4},\frac{5 \pi }{16},\frac{13 \pi }{32},\frac{141 \pi }{256},\frac{399 \pi }{512},\frac{2353 \pi }{2048},\frac{7205 \pi }{4096},\frac{182461 \pi }{65536},\frac{594203 \pi }{131072},\frac{3963051 \pi }{524288},\frac{13477707 \pi }{1048576}\right\}$$

October 20, 2019 04:45 AM

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