Can anyone solve this improper integral?

Find the value of the integral below in terms of kk.

∫k∞tsinh⁡2t dt\int\limits_k^\infty\frac{t}{\sinh^2 t }\,\mathrm dt

#Calculus

Note by A Former Brilliant Member
4 years, 6 months ago

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Comments

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),\sinh(t),\cosh(t),\tanh(t),etc ?

Hint: Use integration by parts along with the results ∫csch2(t) dt=−coth⁡(t)+C\int\textrm{csch}^2(t)\,\mathrm dt=-\coth(t)+C and ∫coth⁡(t) dt=log⁡(sinh⁡(t))+C\int\coth(t)\,\mathrm dt=\log(\sinh(t))+C to find the indefinite integral (say I(t)I(t)). Then, consider the behavior of the hyperbolic functions as t→∞t\to\infty to find the definite integral lim⁡n→∞I(n)−I(k)\lim\limits_{n\to\infty}I(n)-I(k).

Prasun Biswas - 4 years, 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.

A Former Brilliant Member - 4 years, 6 months ago

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

I(n)=−nen+e−nen−e−n+log⁡(en−e−n2)=−n1+e−2n1−e−2n+log⁡(en(1−e−2n))−log⁡2=−n1+e−2n1−e−2n+n+log⁡(1−e−2n)−log⁡2=n(1−1+e−2n1−e−2n)+log⁡(1−e−2n)−log⁡2=−2ne−2n1−e−2n+log⁡(1−e−2n)−log⁡2=−2ne2n−1+log⁡(1−e−2n)−log⁡2\begin{aligned}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{aligned}

Obviously, the second term goes to log⁡(1)=0\log(1)=0 as n→∞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 00 as n→∞n\to\infty, thus giving you the result lim⁡n→∞I(n)=−log⁡(2)\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 −n1+e−2n1−e−2n+n→−n+n=0-n\dfrac{1+e^{-2n}}{1-e^{-2n}}+n\to -n+n=0 as n→∞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 - 4 years, 6 months ago

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@Prasun Biswas – Yeah. Thank you so much again.

A Former Brilliant Member - 4 years, 6 months ago
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