# 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
4 years, 4 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:

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

- 4 years, 4 months ago

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

- 4 years, 4 months ago

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

- 4 years, 4 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 $1$.

- 4 years, 4 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.

- 4 years, 4 months ago

Yes. As $A=1-N-e^x$, so $0 and $0,because we have $-∝, will satisfy the condition.

- 4 years, 4 months ago

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

- 4 years, 4 months ago

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

- 4 years, 4 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.

- 4 years, 4 months ago