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

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

Note by Akash Shukla
1 year, 5 months ago

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The 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 - 1 year, 5 months ago

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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.

Akash Shukla - 1 year, 5 months ago

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

Deeparaj Bhat - 1 year, 5 months ago

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@Deeparaj Bhat I have used the series of \(log\) to solve it.

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

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 - 1 year, 5 months ago

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@Akash Shukla You should have \(-1\leq A <1\)

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 - 1 year, 5 months ago

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@Deeparaj Bhat 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.

Akash Shukla - 1 year, 5 months ago

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@Akash Shukla You didn't integrate properly. If you did, you'd have got infinity a lot of times.

Deeparaj Bhat - 1 year, 5 months ago

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

Akash Shukla - 1 year, 5 months ago

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@Akash Shukla A basic test of convergence is that if \(\lim_{x \to -\infty} f(x) \neq 0 \), then, \(\int_{-\infty}^0 f(x) dx \) diverges.

I think you've not the integration correctly.

Deeparaj Bhat - 1 year, 5 months ago

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