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I-n-0-ln-x-cosh-n-x-dx-is-there-a-simpler-way-to-calculat-those-values-




Question Number 103397 by frc2crc last updated on 14/Jul/20
I(n)=∫_0 ^∞ ((ln x)/(cosh^n x ))dx  is there a simpler way to  calculat those values
$${I}\left({n}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\:{x}}{\mathrm{cosh}^{{n}} {x}\:}{dx} \\ $$$${is}\:{there}\:{a}\:{simpler}\:{way}\:{to} \\ $$$${calculat}\:{those}\:{values} \\ $$
Answered by Aziztisffola last updated on 14/Jul/20
you can use Laplace transform, it′s  the best way to do that, mybe there are  the others.
$$\mathrm{you}\:\mathrm{can}\:\mathrm{use}\:\mathcal{L}\mathrm{aplace}\:\mathrm{transform},\:\mathrm{it}'\mathrm{s} \\ $$$$\mathrm{the}\:\mathrm{best}\:\mathrm{way}\:\mathrm{to}\:\mathrm{do}\:\mathrm{that},\:\mathrm{mybe}\:\mathrm{there}\:\mathrm{are} \\ $$$$\mathrm{the}\:\mathrm{others}. \\ $$
Commented by frc2crc last updated on 14/Jul/20
i couldn′tfind anything
$${i}\:{couldn}'{tfind}\:{anything} \\ $$
Commented by mathmax by abdo last updated on 15/Jul/20
can you take a try sir aziz...
$$\mathrm{can}\:\mathrm{you}\:\mathrm{take}\:\mathrm{a}\:\mathrm{try}\:\mathrm{sir}\:\mathrm{aziz}… \\ $$
Commented by Aziztisffola last updated on 16/Jul/20
I try it but I don′t found the solution  right now .
$$\mathrm{I}\:\mathrm{try}\:\mathrm{it}\:\mathrm{but}\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{found}\:\mathrm{the}\:\mathrm{solution} \\ $$$$\mathrm{right}\:\mathrm{now}\:. \\ $$

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