Quite weird

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

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

#Calculus

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Comments

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:

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

A Former Brilliant Member - 5 years 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.

Akash Shukla - 5 years ago

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

A Former Brilliant Member - 5 years ago

@A Former Brilliant Member – 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 - 5 years ago

@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.

A Former Brilliant Member - 5 years ago

@A Former Brilliant Member – 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 - 5 years ago

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

A Former Brilliant Member - 5 years ago

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

Akash Shukla - 5 years ago

@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.

A Former Brilliant Member - 5 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$