Quite weird

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

Does any value of nn exists such that the above integral becomes 0\Large 0 ?

Note by Akash Shukla
3 years, 4 months ago

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The integral doesn't exist if n1n \neq 1. If the integral exists when n=1n=1, then the integral is positive as the integrand is always positive.

Hence, no value of nn exists such that the integral vanishes.


Edit:

0ln(1+ex)dx=π212\int_{-\infty}^0 \ln (1+e^x) dx = \frac{\pi^2}{12}

Deeparaj Bhat - 3 years, 4 months ago

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I didn't get it. Why the integral doesn't exist for 0<n<10<n<1, as you said the integral exists only for n=1n=1, Thank you.

Akash Shukla - 3 years, 4 months ago

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If n1n \neq 1, then limxlog(n+ex)0\lim_{x \to -\infty} \log (n+e^x) \neq 0 and hence the integral diverges.

Deeparaj Bhat - 3 years, 4 months ago

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@Deeparaj Bhat I have used the series of loglog to solve it.

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

Let, 1nex=A1-n-e^{x} = A,

So, 0log(1A)=A212+A323+A434 ∫_{-∝}^{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 nn which is nearer to 11.

Akash Shukla - 3 years, 4 months ago

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@Akash Shukla You should have 1A<1-1\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 - 3 years, 4 months ago

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@Deeparaj Bhat Yes. As A=1NexA=1-N-e^x, so 0<n<10<n<1 and 0<ex<10<e^x<1,because we have <x<0-∝<x<0, will satisfy the condition.

Akash Shukla - 3 years, 4 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 - 3 years, 4 months ago

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

Akash Shukla - 3 years, 4 months ago

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

I think you've not the integration correctly.

Deeparaj Bhat - 3 years, 4 months ago

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