## klimenkov Why does the formula $\int x^n\,dx=\frac{x^{n+1}}{n+1}+C$give the wrong result if we put $$n=-1$$ ? 11 months ago 11 months ago

1. SandeepReddy

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

2. 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 $$n=-1$$ ?

3. PeterPan

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

4. klimenkov

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

5. PeterPan

Simply, 1/0 does not exist :)

6. klimenkov

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

7. PeterPan

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 $\Large \frac{d}{dx}x^n=nx^{n-1}$

8. PeterPan

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

9. PeterPan

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

10. PeterPan

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

11. klimenkov

Ok. It is time to tell you the truth.

12. PeterPan

I do love the truth :)

13. PeterPan

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

14. klimenkov

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

15. PeterPan

Well, of course, it's still $x^n$

16. klimenkov

But now if you put $$n=-1$$ what will you get?

17. PeterPan

Indeterminate.

18. klimenkov

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

19. PeterPan

0/0

20. klimenkov

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

21. PeterPan

I really don't know.

22. klimenkov

Do you know the L'hopital's rule?

23. PeterPan

No...

24. klimenkov

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

25. klimenkov

@PeterPan Do you know what is Wolfram Alpha?

26. PeterPan

That's why Google is here. :)

27. PeterPan

It seems its limit is ln(x)

28. klimenkov

Yes.

29. PeterPan

30. klimenkov

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

31. PeterPan

Never mind. :)

32. PeterPan

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

33. PeterPan

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...

34. klimenkov

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

35. klimenkov

Now we can include $$n=-1$$ also.

36. PeterPan

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

37. PeterPan

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

38. PeterPan

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

39. klimenkov

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

40. PeterPan

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,

41. 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

42. PeterPan

^^