I have the integral I ( a ) = ∫ 0 ∞ d k k 3 J 1 ( a k ) sinh ⁡ ( k )

Question

I have the integral  I  (  a  )  =      ∫    0    ∞    d  k                       k        3                    J        1            (      a      k      )              sinh      ⁡      (      k      )

conveneau71 · Accepted Answer

Well, we are trying to find the following integral:

\begin{matrix} (1) & I_{n} (α) := \int_{0}^{\infty} \frac{x^{n} J_{1} (α x)}{\sinh (x)} d x = \int_{0}^{\infty} x^{n} J_{1} (α x) csch (x) d x \end{matrix}

Using the 'evaluating integrals over the positive real axis' property of the Laplace transform, we can write:

\begin{matrix} (2) & I_{n} (α) = \int_{0}^{\infty} L_{x} {[x^{n} J_{1} (α x)]}_{(σ)} \cdot L_{x}^{- 1} {[csch (x)]}_{(σ)} d σ \end{matrix}

Using known results, we can write:

\begin{matrix} (3) & I_{n} (α) = {(- 1)}^{n} \cdot \int_{0}^{\infty} \frac{\partial^{n}}{\partial σ^{n}} (\frac{α}{α^{2} + σ (σ + \sqrt{α^{2} + σ^{2}})}) \cdot (2 \sum_{k \geq 0} δ (σ - 2 k - 1)) d σ \end{matrix}

We can rewrite this a bit:

\begin{matrix} (4) & I_{n} (α) = 2 {(- 1)}^{n} \sum_{k \geq 0} \int_{0}^{\infty} \frac{\partial^{n}}{\partial σ^{n}} (\frac{α}{α^{2} + σ (σ + \sqrt{α^{2} + σ^{2}})}) \cdot δ (σ - 2 k - 1) d σ \end{matrix}

Using the property:

\begin{matrix} (5) & \int_{0}^{\infty} y (x) δ (x - p) d x = y (n) θ (n) \end{matrix}

We can write:

\begin{matrix} (6) & I_{n} (α) = 2 {(- 1)}^{n} \sum_{k \geq 0} {θ (1 + 2 k) \cdot {\frac{\partial^{n}}{\partial σ^{n}} (\frac{α}{α^{2} + σ (σ + \sqrt{α^{2} + σ^{2}})}) |}_{σ = 1 + 2 k}} \end{matrix}

Now, using the fact that when

k \in N

we get

θ (1 + 2 k) = 1

. So we can conclude:

\begin{matrix} (7) & I_{n} (α) = 2 {(- 1)}^{n} \sum_{k \geq 0} {{\frac{\partial^{n}}{\partial σ^{n}} (\frac{α}{α^{2} + σ (σ + \sqrt{α^{2} + σ^{2}})}) |}_{σ = 1 + 2 k}} \end{matrix}

I have the integral I ( a ) = <munderover> ∫ 0 <mi mathva

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