Can anyone solve this improper integral?

Find the value of the integral below in terms of \(k\).

\[\int\limits_k^\infty\frac{t}{\sinh^2 t }\,\mathrm dt\]

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1 year, 6 months ago

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What have you tried? Do you know how to compute the indefinite integral of the given integrand? Do you know the standard indefinite integrals of different hyperbolic functions like \(\sinh(t),\cosh(t),\tanh(t),\)etc ?

Hint: Use integration by parts along with the results \(\int\textrm{csch}^2(t)\,\mathrm dt=-\coth(t)+C\) and \(\int\coth(t)\,\mathrm dt=\log(\sinh(t))+C\) to find the indefinite integral (say \(I(t)\)). Then, consider the behavior of the hyperbolic functions as \(t\to\infty\) to find the definite integral \(\lim\limits_{n\to\infty}I(n)-I(k)\).

Prasun Biswas - 1 year, 6 months ago

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Thank you so much for the beautiful response. Coming to problem, actually I did the same. But I have ended up with two functions whose limit as 't' tends to infinity is again become a question. They are (t coth (t)) and log (sinh (t)) . They are going to infinity. I'm thinking that, this integrand will not converge at all.

Brilliant Member - 1 year, 6 months ago

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Their individual limits do diverge but their difference (which is \(I(n)\), does converge). By now, I assume you must have found out that we have \(I(n)=-n\coth(n)+\log(\sinh(n))\). Using the definitions of the hyperbolic functions, we have,

\[\begin{align}I(n)=-n\frac{e^n+e^{-n}}{e^n-e^{-n}}+\log\left(\frac{e^n-e^{-n}}{2}\right)&=-n\frac{1+e^{-2n}}{1-e^{-2n}}+\log(e^n(1-e^{-2n}))-\log 2\\&=-n\frac{1+e^{-2n}}{1-e^{-2n}}+n+\log(1-e^{-2n})-\log 2\\&=n\left(1-\frac{1+e^{-2n}}{1-e^{-2n}}\right)+\log(1-e^{-2n})-\log 2\\&=\frac{-2ne^{-2n}}{1-e^{-2n}}+\log(1-e^{-2n})-\log 2\\&=\frac{-2n}{e^{2n}-1}+\log(1-e^{-2n})-\log 2\end{align}\]

Obviously, the second term goes to \(\log(1)=0\) as \(n\to\infty\). For the first term, note that it's an indeterminate form, so using L'Hopital's rule will show that the first term also goes to \(0\) as \(n\to\infty\), thus giving you the result \(\lim\limits_{n\to\infty}I(n)=-\log(2)\).


You could also informally argue that since the exponential function increases/decreases rapidly than a polynomial function, we have \(-n\dfrac{1+e^{-2n}}{1-e^{-2n}}+n\to -n+n=0\) as \(n\to\infty\) which would provide you the same final result but I think the above approach is more formal and rigorous.

Can you do the rest now?

Prasun Biswas - 1 year, 6 months ago

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@Prasun Biswas Yeah. Thank you so much again.

Brilliant Member - 1 year, 6 months ago

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