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# Quite weird

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

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:

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

- 1 year, 5 months 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.

- 1 year, 5 months ago

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

- 1 year, 5 months ago

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 $$1$$.

- 1 year, 5 months ago

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.

- 1 year, 5 months 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.

- 1 year, 5 months ago

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

- 1 year, 5 months ago

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

- 1 year, 5 months ago

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.

- 1 year, 5 months ago