Converting a limit to integral form or vice-versa

In summary: You can't apply L'Hopital's rule here because the numerator goes to 0 while the denominator goes to -infinity. So you end up with an indeterminate form of 0/∞. You need to use a different method to evaluate this limit, such as substitution or series expansion.In summary, the proof for the given integral is based on the relationship between integrals and Riemann sums. By taking the limit of a summation with infinitely small rectangles, trapezoids, or other shapes, we can find the area under a curve. This can be written as a definite integral, and by setting x0 = a, x1 = a + h, etc., we can derive the given formula. However
  • #1
Raghav Gupta
1,011
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What is the proof for this
$$ \int_a^b f(x) dx = 1/n\lim_{n\to\infty} (f(a) + f(a+h) + f(a+2h) +...+ f( a+ (n-1)h)) $$
h = (b-a)/n

Also I think there is some summation form which can be converted to integral form how?
 
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  • #2
What is your definition of the integral?
 
  • #3
  • #4
micromass said:
What is your definition of the integral?
BvU said:
Hi Raghav,

For this you can read up under Riemann sum

As μmass already indicates: an integral and a Riemann sum are closely related.

See also Riemann integral , Fundamental theorem of calculus
An integral is the area under a curve.
For any curve we can find it's area by putting infinitely small rectangles, trapezoids etc. shapes.
We can write $$ \int_a^b f(x) dx = \lim_{n\to\infty} \sum_{i=1}^n f(x_{i-1}) Δx $$
Δx = (b-a)/n
But how do we get this

$$ \int_a^b f(x) dx = 1/n\lim_{n\to\infty} (f(a) + f(a+h) + f(a+2h) +...+ f( a+ (n-1)h)) $$
h = (b-a)/n
From above that I know?
 
  • #5
By setting x0 = a, x1 = a + h, etc. (you already have ##\Delta x = h##) .

I think there is a small error in the lower formula: 1/n can't be outside the ##\lim## .
(It also doesn't fit dimensionwise)
(b-a) should be the factor in front and 1/n should be at the end (within the ##\lim## scope)
 
  • #6
BvU said:
By setting x0 = a, x1 = a + h, etc. (you already have ##\Delta x = h##) .

I think there is a small error in the lower formula: 1/n can't be outside the ##\lim## .
(It also doesn't fit dimensionwise)
(b-a) should be the factor in front and 1/n should be at the end (within the ##\lim## scope)
Okay, got that and yes 1/n should be inside lim scope.

Applying this how $$ \lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^n log(r/n) = \int_0^1 logx dx$$ ?
 
  • #7
Ah ! A new turn to this thread. We're going to derive Stirling's formula !

Your righthand side makes me feel uncomfortable. A primitive of ##\log x## is ##x\log x - x## and there is no ##\log 0##, so I suppose this is to be continued. And yes:

If you take ##f(x) = \log x## and write out
$$
\int_a^b f(x) dx = (b-a) \lim_{n\to\infty} ( f(a+h) + f(a+2h) +...+ f( a+ nh)) 1/n
$$you see your summation appearing (b = 1, a = 0, h = 1/n).

Note that I sneakily shifted by h (starting from a+h instead of from a). So from lower sum to upper sum (which google) to avoid the ##\log 0##).​
 
  • #8
BvU said:
Ah ! A new turn to this thread. We're going to derive Stirling's formula !

Your righthand side makes me feel uncomfortable. A primitive of ##\log x## is ##x\log x - x## and there is no ##\log 0##, so I suppose this is to be continued. And yes:

If you take ##f(x) = \log x## and write out
$$
\int_a^b f(x) dx = (b-a) \lim_{n\to\infty} ( f(a+h) + f(a+2h) +...+ f( a+ nh)) 1/n
$$you see your summation appearing (b = 1, a = 0, h = 1/n).

Note that I sneakily shifted by h (starting from a+h instead of from a). So from lower sum to upper sum (which google) to avoid the ##\log 0##).​
I saw that Stirling formula in Wikipedia and I am feeling uncomfortable with that. ( I think it is taught in university with a lot of basics first and I am in high school.)
I also googled lower to upper sum and don't know that things trapezoidal rule in details.( I think university will be fun.:rolleyes:)

So, according to wolframalpha
Left hand side is 0 and Right hand side is coming -1
Is that true?
But someone was telling me that they are equal.
Also 1/∞ is not strictly zero, it may be 0.00000...1
So log 0 not comes in left hand side.
 
  • #9
Raghav Gupta said:
Also 1/∞ is not strictly zero, it may be 0.00000...1

##1/\infty## is undefined. It doesn't exist. And if you do want to define it, then it will be exactly 0.
 
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Likes Raghav Gupta
  • #10
micromass said:
##1/\infty## is undefined. It doesn't exist. And if you do want to define it, then it will be exactly 0.
Okay, so is this true?

$$ \lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^n log(r/n) = \int_0^1 logx dx$$

[Edit - added ] is 0 * log0 = 0 ?
 
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  • #11
is 0 * log0 = 0
No, but ##\displaystyle \lim_{x\rightarrow 0} x\log x = 0##

Left hand side is 0 and Right hand side is coming -1
Where do you see a zero appearing on the left hand side ? Try it on your calculator (mine ?:) goes up to n = 170 and no way it's going to 0).

So is Wolfram a dud ? No it's not. But you need to improve the incantation...

By the way, my deepest respect for the JEE level. I find it hard to believe they can ask so much of people wanting to get into university ! All (or almost) all the stuff I have to mobilize in order to help you, I learned after grammar school.​
 
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  • #12
BvU said:
No, but ##\displaystyle \lim_{x\rightarrow 0} x\log x = 0##

Where do you see a zero appearing on the left hand side ? Try it on your calculator (mine ?:) goes up to n = 170 and no way it's going to 0).

So is Wolfram a dud ? No it's not. But you need to improve the incantation...
​
##\displaystyle \lim_{x\rightarrow 0} x\log x = 0##
Now how is this true?

Got the -1 in left hand side by improving incantation:)
 
  • #13
As x goes to 0, log(x) along goes to -infinity so [itex]\frac{1}{log(x)}[/itex] goes to 0 and we can apply L'Hopital's rule to [itex]\frac{x}{\frac{1}{log(x)}}[/itex].

The derivative of x is, or course, 1 and the derivative of [itex]\frac{1}{log(x)}= (log(x))^{-1}[/itex] is [itex]-(log(x))^{-2}\frac{1}{x}[/itex] so that the limit is the limit of [itex]\frac{1}{-\frac{1}{x}(log(x))^2}[/itex][itex]= -\frac{(log(x))^2}{x}= -\frac{xlog(x)}{x}[/itex].

So if [itex]\lim_{x\to 0} x log(x)[/itex] is some finite number, A, then we must have [itex]A= -A(\lim_{x\to 0} x)= 0[/itex]
 
  • #14
HallsofIvy said:
As x goes to 0, log(x) along goes to -infinity so [itex]\frac{1}{log(x)}[/itex] goes to 0 and we can apply L'Hopital's rule to [itex]\frac{x}{\frac{1}{log(x)}}[/itex].

The derivative of x is, or course, 1 and the derivative of [itex]\frac{1}{log(x)}= (log(x))^{-1}[/itex] is [itex]-(log(x))^{-2}\frac{1}{x}[/itex] so that the limit is the limit of [itex]\frac{1}{-\frac{1}{x}(log(x))^2}[/itex][itex]= -\frac{(log(x))^2}{x}= -\frac{xlog(x)}{x}[/itex].

So if [itex]\lim_{x\to 0} x log(x)[/itex] is some finite number, A, then we must have [itex]A= -A(\lim_{x\to 0} x)= 0[/itex]
Shouldn't that bold part have## (log(x))^{-2} ## in denominator ?
and I am also not getting what you are doing after that
How
[itex]\frac{1}{-\frac{1}{x}(log(x))^2}[/itex][itex]= -\frac{(log(x))^2}{x}[/itex]
 
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  • #15
  • #16
HaHa, no problem,
Halls you took me in a complicated way,
Here see this,
$$ xlogx = \frac{logx}{\frac{1}{x}} $$
Here limit x tends to zero and as we are getting num. and den -∞/∞ we can then also apply L Hopital rule.
Then
$$ \frac{-x^2}{x} = -x $$
As here limit x tends to 0
We get 0.

Thanks BvU, Halls (although your way was very complicated but bringing L Hopital was crucial) and also thanks to μm .
 

Related to Converting a limit to integral form or vice-versa

1. How do I convert a limit to integral form?

The process of converting a limit to integral form involves taking the limit of a function as the independent variable approaches a specific value, and then rewriting it as an integral with that variable as the upper or lower limit. This can be done by using the fundamental theorem of calculus and the definition of a limit.

2. Why would I want to convert a limit to integral form?

Converting a limit to integral form can be useful in solving certain types of problems in calculus, such as finding the area under a curve or calculating the volume of a solid. It can also help to simplify calculations and make the problem easier to solve.

3. Can any limit be converted to integral form?

No, not all limits can be converted to integral form. The function must be continuous and the limit must exist at the value being approached in order for it to be converted. Additionally, the limit must be able to be written as an integral using the fundamental theorem of calculus.

4. How do I convert an integral to limit form?

The process of converting an integral to limit form involves using the fundamental theorem of calculus to evaluate the integral and then taking the limit as the upper or lower limit approaches a specific value. This can be done by rewriting the integral as a limit using the definition of a limit.

5. Are there any limitations to converting between limit and integral form?

One limitation is that the function being evaluated must be continuous and the limit must exist at the value being approached. Additionally, the integral must be able to be evaluated using the fundamental theorem of calculus. Another limitation is that this method may not always be the most efficient or accurate way to solve a problem, so it is important to consider other methods as well.

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