A Complex Integral
Let η(x) represent the (√(-1) + 1)th integral of the hyperbolic sine function, with any arbitrary constants of integration equaling zero. Express η(π) to four decimal places, rounding if necessary.
Note: Assume that η(x) is the particular solution where 0 < θ < 2π on the complex unit circle.
The answer is 12.0703.
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1 solution
-Note: The nth integral of f(x) is equal to the (-n)th derivative of f(x)
-Note: The hyperbolic sine function may be rewritten as (1/2)[(e^x)-(e^-x)]
-Note: The first derivative of sinh(x) = cosh(x) = (1/2)[(e^x)+(e^-x)]
Through successive differentiation, one may observe that d^n/dx^n[sinh(x)] may be generalised through the expression (1/2)[(e^x)-((-1)^n)(e^-x)] = (1/2)[(e^x)-(e^(π𝑖n-x)], the latter expression of which adheres to initial constraints, as 0 < π < 2π
Plug in -(𝑖+1) for n to find that η(x) = (1/2)[(e^x)+(e^(π-x))]
Thus η(π) = (1/2)((e^π)+1) ≈ 12.0703