step by step solution of integral 1/sqrt(1-u^4)

Question

How to solve this integral to obtain its value within the interval 0 to 1.

Answer 1

∫du/√(1-u⁴), u∈[0,1]. This definite integral must tend to infinity because 1/√(1-u⁴) is not defined when u=1. The integral is the area beneath the curve between the given limits. u=1 is an asymptote so this area is infinite.

Series method:

1/√(1-u⁴)=(1-u⁴)^-½=1+u⁴/2+(1)(3)u⁸/2²+(1)(3)(5)u¹²/2³+...+[(2n-1)(2n-3)...(3)(1)u⁴ⁿ/2ⁿ]+... 

We can write the series as ∑(2n-1)(2n-3)...(3)(1)u⁴ⁿ/2ⁿ for n≥0. (n!=1 for n≤0)

This can be written ∑[(2n-1)!/(2^n-1(n-1)!)]u⁴ⁿ/2ⁿ=∑(2n-1)!u⁴ⁿ/[2^2n-1(n-1)!]

Example: the 4th term (n=3) is (1)(3)(5)u¹²/2³=15u¹²/8; using the formula: 5!u¹²/[2⁵(2!)]=120u¹²/64=15u¹²/8.

So ∫∑(2n-1)!u⁴ⁿ/[2^2n-1(n-1)!]du=∑(2n-1)!u⁴ⁿ⁺¹/[2^2n-1(n-1)!(4n+1)].

Example: n=3: 120u¹³/(64×13)=15u¹³/104 is the 4th term after integration.

Applying the limits we get:

∑(2n-1)!/[2^2n-1(n-1)!(4n+1)]=

1+1/10+1/12+15/104+105/272+45/32+2079/320+... which tends to infinity, because the terms get larger as the series progresses.