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Question Number 146090 by Ar Brandon last updated on 10/Jul/21

$${\int }_0^{\mathrm{\infty }}\frac{\sinh \left(\mathrm{at}\right)\sinh \left(\mathrm{bt}\right)}{\sinh \left(\mathrm{ct}\right){\mathrm{e}}^{\mathrm{tz}}}\mathrm{dt}=$$$$\frac{\mathrm{ab}}{\mathrm{c}({\mathrm{z}}^2+{\mathrm{c}}^2-{\mathrm{a}}^2-{\mathrm{b}}^2+\underset{\mathrm{k}=1}{\stackrel{\mathrm{\infty }}{\mathrm{K}}}\frac{-{4\mathrm{k}}^2({\mathrm{k}}^2{\mathrm{c}}^2-{\mathrm{a}}^2)({\mathrm{k}}^2{\mathrm{c}}^2-{\mathrm{b}}^2)}{(2\mathrm{k}+1)({\mathrm{z}}^2+({2\mathrm{k}}^2+2\mathrm{k}+1){\mathrm{c}}^2-{\mathrm{a}}^2-{\mathrm{b}}^2)})}$$

Answered by mindispower last updated on 10/Jul/21

$$zisx?$$

Commented by Ar Brandon last updated on 10/Jul/21

$$\mathrm{Thanks}\mathrm{for}\mathrm{the}\mathrm{remark}.\mathrm{I}\mathrm{appologize}\mathrm{for}\mathrm{that}.$$$$\mathrm{It}\mathrm{was}\mathrm{a}\mathrm{mistake},{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{been}\mathrm{edited}.$$

Answered by mindispower last updated on 14/Jul/21

$$tchekitsir$$

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