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There is an integral representation for the modified Bessel function of the second (or third depending on who you talk to) kind (denoted $K_\nu$) that says:

$$K_\nu(z) = \dfrac{\sqrt{\pi}(\frac{1}{2}z)^\nu}{\Gamma(\nu+\frac{1}{2})}\int_1^\infty e^{-z t}(t^2-1)^{\nu-\frac{1}{2}}dt,$$ where $\Re(\nu)>-\frac{1}{2}$, and $|\mathrm{Arg}(z)|<\frac{\pi}{2}$.

My question is this: if we limit ourselves to the conditions that $\nu\in\mathbb{R}$ and $z\in\mathbb{R}$, we always have order symmetry, i.e. $$K_{-\nu}(z)=K_\nu(z).$$ But this equation does not appear to be symmetric with respect to $\nu$, namely putting $\pm\frac{1}{4}$ would appear to give different equations. Am I missing something?

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It is known that the modified Bessel Function $K_z(a)$ ($a>0$)can be expressed as a Fourier transform $$K_\nu(z)=\frac{1}{2}\int_{-\infty}^{\infty}\exp(-z\cosh u)\cosh(\nu u){\rm d}u=K_{-\nu}(z)$$

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