# How to compute $I_{m,n,k}=\int_{0}^{\infty} \frac{dx}{\sinh^n(x) \coth^{2k+n+m-1}(x)}$?

To compute $$I_{m,n,k}$$, expressing the integrand function with the function $$\tanh$$. As $$\coth(x)=\frac{\cosh(x)}{\sinh(x)}$$ and $$\frac{1}{\cosh^2(x)}= 1 -\tanh^2(x)$$, we get \begin{align} I_{m,n,k}&=\int_{0}^{\infty} \frac{dx}{\sinh^n(x) \coth^{2k+n+m-1}(x)} \\ &=\int_{0}^{\infty} \frac{\tanh^{2k+m-1}(x)}{(\sinh(x)\coth(x))^{n}} dx\\ &=\int_{0}^{\infty} \frac{\tanh^{2k+m-1}(x)}{\cosh^{n}(x)} dx\\ &=\int_{0}^{\infty} \tanh^{2k+m-1}(x) (1-\tanh^{2}(x))^{\frac{n}{2}}dx\\ &= ....? \end{align} Otherwise, as $$\tanh(t)'=1 -\tanh(t)^2$$, we get \begin{align} I_{m,n,k}&=\int_{0}^{\infty} \tanh^{2k+m-1}(x) (\tanh(x)')^{\frac{n}{2}}dx= ?? \end{align}

I don't know how to proceed. Any help is appreciated. Thanks in advance.

Why don't you write $$\int_0^\infty\tanh^{2k+m-1}x(1-\tanh^2x)^{\frac n2 -1}(1-\tanh^2x) dx$$ Then you do the change of variable $$y=\tanh x$$, and you get $$\int_0^1y^{2k+m-1}(1-y^2)^{\frac n2 -1} dy$$