Why does the formula \[\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\]give the wrong result if we put \(n=-1\) ?

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- klimenkov

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- anonymous

AS you only stated the formula, substituting n = -1 in the formula on RHS, gives undefined form

- klimenkov

This not the answer I want to hear.
Why does this formula give the right result if we put any \(n\) except \(n=-1\) ? May be there is any trick for \(n=-1\) ?

- anonymous

Because the antiderivative of x^{-1} is ln(x)

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## More answers

- klimenkov

\[\int \frac1x\,dx=\frac{x^0}{0}+C=\frac10+C\]What is wrong with it?

- anonymous

Simply, 1/0 does not exist :)

- klimenkov

\[\int \frac1x\,dx=\frac{x^0}{0}+C=\frac10+C=\ln x+C\]\[\frac10=\ln x\]

- anonymous

Maybe it's best to recall that we have this formula
\[\int x^n\,dx=\frac{x^{n+1}}{n+1}+C\]BECAUSE of this formula
\[\Large \frac{d}{dx}x^n=nx^{n-1}\]

- anonymous

Or, more relevantly
\[\Large \frac{d}{dx}\frac{x^{n+1}}{n+1}=x^n\]

- anonymous

But you see that it no longer applies for n = -1

- anonymous

Because
\[\Large \frac{d}{dx}\frac{x^{-1+1}}{-1+1}=\frac{d}{dx}\frac10\]and 1/0 never existed to begin with :/

- klimenkov

Ok. It is time to tell you the truth.

- anonymous

I do love the truth :)

- anonymous

Noting of course, that there are many ways to explain this.

- klimenkov

Try to compute\[\frac{d}{dx}\left(\frac{x^{n+1}-1}{n+1}\right)\]

- anonymous

Well, of course, it's still
\[x^n\]

- klimenkov

But now if you put \(n=-1\) what will you get?

- anonymous

Indeterminate.

- klimenkov

Don't mind on indeterminate. Just compute this like a simple limit.

- anonymous

0/0

- klimenkov

\(\frac{\sin n}{n}\) is \(\frac00\) too, but we all know that it is equal to 1. What is that limit?

- anonymous

I really don't know.

- klimenkov

Do you know the L'hopital's rule?

- anonymous

No...

- klimenkov

\[\lim_{n\rightarrow-1}\frac{x^{n+1}-1}{n+1}\]Anybody here know how to compute this limit? @ParthKohli

- klimenkov

@PeterPan
Do you know what is Wolfram Alpha?

- anonymous

That's why Google is here. :)

- anonymous

It seems its limit is ln(x)

- klimenkov

Yes.

- anonymous

Your point?

- klimenkov

Sorry, cant understand what means "Your point?".

- anonymous

Never mind. :)

- anonymous

You did say you were looking for a specific answer, so there was preference :P

- anonymous

Besides, you gave a specific example, which is not valid for proofs :)
It has to be true for all constants. I think I'll stand by my original statements, interesting though this was...

- klimenkov

Actually, the right expression is
\[\int x^n\,dx=\frac{x^{n+1}-1}{n+1}+C\]

- klimenkov

Now we can include \(n=-1\) also.

- anonymous

No you can't. It has to be
\[\Large \int x^k\,dx=\lim_{k\rightarrow n}\frac{x^{n+1}-1}{n+1}+C\]

- anonymous

\[\Large \int x^k\,dx=\lim_{n\rightarrow k}\frac{x^{n+1}-1}{n+1}+C\]

- anonymous

Without the limit, it still won't make sense.

- klimenkov

Very nice. But forget using limits. It is restrictions that really does not play a big role.

- anonymous

Finally. But I find it easier to just affirm that the modified power rule
\[\Large \frac{d}{dx}\frac{x^{n+1}}{n+1}=x^n\]does not hold for n=-1
And just proving that the derivative of ln(x) is 1/x,

- shubhamsrg

the diff is there since 1/x is not continuous at x=0
x^n always passes through (0,0) if it is not equal to -1, but for n=-1, the shape of graph is completely different. And, integration is simply the area under the graph.
I know my logic is very wayward and not very strong, but still.. a point worth mentioning. :P

- anonymous

^^

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