I have the integral

$I(a)=\underset{0}{\overset{\mathrm{\infty}}{\int}}dk\text{}\frac{{k}^{3}{J}_{1}(ak)}{\mathrm{sinh}(k)}$

$I(a)=\underset{0}{\overset{\mathrm{\infty}}{\int}}dk\text{}\frac{{k}^{3}{J}_{1}(ak)}{\mathrm{sinh}(k)}$

Aganippe76
2022-07-06
Answered

I have the integral

$I(a)=\underset{0}{\overset{\mathrm{\infty}}{\int}}dk\text{}\frac{{k}^{3}{J}_{1}(ak)}{\mathrm{sinh}(k)}$

$I(a)=\underset{0}{\overset{\mathrm{\infty}}{\int}}dk\text{}\frac{{k}^{3}{J}_{1}(ak)}{\mathrm{sinh}(k)}$

You can still ask an expert for help

conveneau71

Answered 2022-07-07
Author has **17** answers

Well, we are trying to find the following integral:

$\begin{array}{}\text{(1)}& {\mathcal{I}}_{\text{n}}\left(\alpha \right):={\int}_{0}^{\mathrm{\infty}}\frac{{x}^{\text{n}}{\mathcal{J}}_{1}\left(\alpha x\right)}{\mathrm{sinh}\left(x\right)}\text{}\text{d}x={\int}_{0}^{\mathrm{\infty}}{x}^{\text{n}}{\mathcal{J}}_{1}\left(\alpha x\right)\text{csch}\left(x\right)\text{}\text{d}x\end{array}$

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

$\begin{array}{}\text{(2)}& {\mathcal{I}}_{\text{n}}\left(\alpha \right)={\int}_{0}^{\mathrm{\infty}}{\mathcal{L}}_{x}{\left[{x}^{\text{n}}{\mathcal{J}}_{1}\left(\alpha x\right)\right]}_{\left(\sigma \right)}\cdot {\mathcal{L}}_{x}^{-1}{\left[\text{csch}\left(x\right)\right]}_{\left(\sigma \right)}\text{}\text{d}\sigma \end{array}$

Using known results, we can write:

$\begin{array}{}\text{(3)}& {\mathcal{I}}_{\text{n}}\left(\alpha \right)={(-1)}^{\text{n}}\cdot {\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{\partial}}^{\text{n}}}{\mathrm{\partial}{\sigma}^{\text{n}}}\left(\frac{\alpha}{{\alpha}^{2}+\sigma (\sigma +\sqrt{{\alpha}^{2}+{\sigma}^{2}})}\right)\cdot \left(2\sum _{\text{k}\text{}\ge \text{}0}\delta (\sigma -2\text{k}-1)\right)\text{}\text{d}\sigma \end{array}$

We can rewrite this a bit:

$\begin{array}{}\text{(4)}& {\mathcal{I}}_{\text{n}}\left(\alpha \right)=2{(-1)}^{\text{n}}\sum _{\text{k}\text{}\ge \text{}0}{\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{\partial}}^{\text{n}}}{\mathrm{\partial}{\sigma}^{\text{n}}}\left(\frac{\alpha}{{\alpha}^{2}+\sigma (\sigma +\sqrt{{\alpha}^{2}+{\sigma}^{2}})}\right)\cdot \delta (\sigma -2\text{k}-1)\text{}\text{d}\sigma \end{array}$

Using the property:

$\begin{array}{}\text{(5)}& {\int}_{0}^{\mathrm{\infty}}\text{y}\left(x\right)\delta (x-\text{p})\text{}\text{d}x=\text{y}\left(\text{n}\right)\theta \left(\text{n}\right)\end{array}$

We can write:

$\begin{array}{}\text{(6)}& {\mathcal{I}}_{\text{n}}\left(\alpha \right)=2{(-1)}^{\text{n}}\sum _{\text{k}\text{}\ge \text{}0}\{\theta (1+2\text{k})\cdot {\frac{{\mathrm{\partial}}^{\text{n}}}{\mathrm{\partial}{\sigma}^{\text{n}}}\left(\frac{\alpha}{{\alpha}^{2}+\sigma (\sigma +\sqrt{{\alpha}^{2}+{\sigma}^{2}})}\right)|}_{\text{}\sigma \text{}=\text{}1+2\text{k}}\}\end{array}$

Now, using the fact that when $\text{k}\in \mathbb{N}$ we get $\theta (1+2\text{k})=1$. So we can conclude:

$\begin{array}{}\text{(7)}& {\mathcal{I}}_{\text{n}}\left(\alpha \right)=2{(-1)}^{\text{n}}\sum _{\text{k}\text{}\ge \text{}0}\left\{{\frac{{\mathrm{\partial}}^{\text{n}}}{\mathrm{\partial}{\sigma}^{\text{n}}}\left(\frac{\alpha}{{\alpha}^{2}+\sigma (\sigma +\sqrt{{\alpha}^{2}+{\sigma}^{2}})}\right)|}_{\text{}\sigma \text{}=\text{}1+2\text{k}}\right\}\end{array}$

$\begin{array}{}\text{(1)}& {\mathcal{I}}_{\text{n}}\left(\alpha \right):={\int}_{0}^{\mathrm{\infty}}\frac{{x}^{\text{n}}{\mathcal{J}}_{1}\left(\alpha x\right)}{\mathrm{sinh}\left(x\right)}\text{}\text{d}x={\int}_{0}^{\mathrm{\infty}}{x}^{\text{n}}{\mathcal{J}}_{1}\left(\alpha x\right)\text{csch}\left(x\right)\text{}\text{d}x\end{array}$

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

$\begin{array}{}\text{(2)}& {\mathcal{I}}_{\text{n}}\left(\alpha \right)={\int}_{0}^{\mathrm{\infty}}{\mathcal{L}}_{x}{\left[{x}^{\text{n}}{\mathcal{J}}_{1}\left(\alpha x\right)\right]}_{\left(\sigma \right)}\cdot {\mathcal{L}}_{x}^{-1}{\left[\text{csch}\left(x\right)\right]}_{\left(\sigma \right)}\text{}\text{d}\sigma \end{array}$

Using known results, we can write:

$\begin{array}{}\text{(3)}& {\mathcal{I}}_{\text{n}}\left(\alpha \right)={(-1)}^{\text{n}}\cdot {\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{\partial}}^{\text{n}}}{\mathrm{\partial}{\sigma}^{\text{n}}}\left(\frac{\alpha}{{\alpha}^{2}+\sigma (\sigma +\sqrt{{\alpha}^{2}+{\sigma}^{2}})}\right)\cdot \left(2\sum _{\text{k}\text{}\ge \text{}0}\delta (\sigma -2\text{k}-1)\right)\text{}\text{d}\sigma \end{array}$

We can rewrite this a bit:

$\begin{array}{}\text{(4)}& {\mathcal{I}}_{\text{n}}\left(\alpha \right)=2{(-1)}^{\text{n}}\sum _{\text{k}\text{}\ge \text{}0}{\int}_{0}^{\mathrm{\infty}}\frac{{\mathrm{\partial}}^{\text{n}}}{\mathrm{\partial}{\sigma}^{\text{n}}}\left(\frac{\alpha}{{\alpha}^{2}+\sigma (\sigma +\sqrt{{\alpha}^{2}+{\sigma}^{2}})}\right)\cdot \delta (\sigma -2\text{k}-1)\text{}\text{d}\sigma \end{array}$

Using the property:

$\begin{array}{}\text{(5)}& {\int}_{0}^{\mathrm{\infty}}\text{y}\left(x\right)\delta (x-\text{p})\text{}\text{d}x=\text{y}\left(\text{n}\right)\theta \left(\text{n}\right)\end{array}$

We can write:

$\begin{array}{}\text{(6)}& {\mathcal{I}}_{\text{n}}\left(\alpha \right)=2{(-1)}^{\text{n}}\sum _{\text{k}\text{}\ge \text{}0}\{\theta (1+2\text{k})\cdot {\frac{{\mathrm{\partial}}^{\text{n}}}{\mathrm{\partial}{\sigma}^{\text{n}}}\left(\frac{\alpha}{{\alpha}^{2}+\sigma (\sigma +\sqrt{{\alpha}^{2}+{\sigma}^{2}})}\right)|}_{\text{}\sigma \text{}=\text{}1+2\text{k}}\}\end{array}$

Now, using the fact that when $\text{k}\in \mathbb{N}$ we get $\theta (1+2\text{k})=1$. So we can conclude:

$\begin{array}{}\text{(7)}& {\mathcal{I}}_{\text{n}}\left(\alpha \right)=2{(-1)}^{\text{n}}\sum _{\text{k}\text{}\ge \text{}0}\left\{{\frac{{\mathrm{\partial}}^{\text{n}}}{\mathrm{\partial}{\sigma}^{\text{n}}}\left(\frac{\alpha}{{\alpha}^{2}+\sigma (\sigma +\sqrt{{\alpha}^{2}+{\sigma}^{2}})}\right)|}_{\text{}\sigma \text{}=\text{}1+2\text{k}}\right\}\end{array}$

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