New user? Sign up

Existing user? Log in

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

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

Note by Akash Shukla 2 years, 10 months ago

Easy Math Editor

*italics*

_italics_

**bold**

__bold__

- bulleted- list

1. numbered2. list

paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)

> This is a quote

This is a quote

# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

Sort by:

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:

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

Log in to reply

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.

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

@Deeparaj Bhat – 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 – 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.

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

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

@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.

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

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.

Log in to reply

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

Log in to reply

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

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 \(1\).

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

Log in to reply

I think you've not the integration correctly.

Log in to reply