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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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

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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.

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

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$log$ to solve it.

I have used the series of$log(n+x) = log[1-(1-n-e^{x})]$,

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 $1$.

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$-1\leq A <1$

You should havePlus, you didn't integrate in the right way. If you did, you'll end up with infinty right from the first term.

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

Yes. AsLog in to reply

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$n<1, n>0$ . Is there any mistake in the method I have shown?

I got. But not for values ofLog in to reply

$\lim_{x \to -\infty} f(x) \neq 0$, then, $\int_{-\infty}^0 f(x) dx$ diverges.

A basic test of convergence is that ifI think you've not the integration correctly.

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