\[\Large ∫_{-∝}^{0} log (n+e^{x})\] \(n\in R\)

Does any value of \(n\) exists such that the above integral becomes \(\Large 0\) ?

\[\Large ∫_{-∝}^{0} log (n+e^{x})\] \(n\in R\)

Does any value of \(n\) exists such that the above integral becomes \(\Large 0\) ?

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TopNewestThe integral doesn't exist if \(n \neq 1\). If the integral exists when \(n=1\), then the integral is positive as the integrand is always positive.

Hence, no value of \(n\) exists such that the integral vanishes.

Edit:\[\int_{-\infty}^0 \ln (1+e^x) dx = \frac{\pi^2}{12} \] – Deeparaj Bhat · 9 months, 3 weeks ago

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– Akash Shukla · 9 months, 3 weeks ago

I didn't get it. Why the integral doesn't exist for \(0<n<1\), as you said the integral exists only for \(n=1\), Thank you.Log in to reply

– Deeparaj Bhat · 9 months, 3 weeks ago

If \(n \neq 1\), then \(\lim_{x \to -\infty} \log (n+e^x) \neq 0\) and hence the integral diverges.Log in to reply

\(log(n+x) = log[1-(1-n-e^{x})] \),

Let, \(1-n-e^{x} = A\),

So, \( ∫_{-∝}^{0} log (1-A) = \dfrac{A^2}{1*2}+\dfrac{A^3}{2*3}+\dfrac{A^4}{3*4}\cdots \),

Here it can be seen that it will converge for some values of \(n\) which is nearer to \(1\). – Akash Shukla · 9 months, 3 weeks ago

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Plus, you didn't integrate in the right way. If you did, you'll end up with infinty right from the first term. – Deeparaj Bhat · 9 months, 3 weeks ago

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– Akash Shukla · 9 months, 3 weeks ago

Yes. As \(A=1-N-e^x\), so \(0<n<1\) and \(0<e^x<1\),because we have \(-∝<x<0\), will satisfy the condition.Log in to reply

– Deeparaj Bhat · 9 months, 3 weeks ago

You didn't integrate properly. If you did, you'd have got infinity a lot of times.Log in to reply

– Akash Shukla · 9 months, 3 weeks ago

I got. But not for values of \(n<1, n>0\) . Is there any mistake in the method I have shown?Log in to reply

I think you've not the integration correctly. – Deeparaj Bhat · 9 months, 3 weeks ago

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