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# Can anyone solve this improper integral?

Find the value of the integral below in terms of $$k$$.

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

Note by Brilliant Member
1 year, 1 month 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 $$\lim\limits_{n\to\infty}I(n)-I(k)$$.

- 1 year, 1 month ago

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.

- 1 year, 1 month ago

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?

- 1 year, 1 month ago

Yeah. Thank you so much again.

- 1 year, 1 month ago