@Amir,

please follow the advice of Issam and give your formula explicitely!

I can't see how anyone can expect to get a useful answer to such a carelessly formulated question.

What function is it precisely? What is the meaning of those zeroes? Some limit of the type "zero divided by zero"?

Yes try L'Hospital rule to find the limit of such a function.

Without knowing the exact integral, it is difficult to know the interpretation of your results. We cannot know if the hypothesis needed by the l'Hospital "rule" are satisfied, particularly that the function must be differentiable on an open interval around the limit toward which you want to compute it!

https://en.wikipedia.org/wiki/L%27H%C3%B4pital%27s_rule

Even if it applies, it is not clear that you could commute the derivative and the integral to simplify your computations...

@Amir Jamei Osgouei

My suggestion is that before putting the both the limits apply L Hospital rules then use the limits. Hope your problems will be solved .

B.Rath

In alternative, you can use alternative approaches as the residue theorem of the complex analysis. Some examples are also in Wikipedia.

Dear Amir,

Just write the integral or show it in a pdf file, then we show you the exact answer.

It seems you have a problem with (1) the boundaries of the integral or

(2) the integrand function is not defined over the interval of integration.

That makes the integral is an improper one. It may diverge or converge.

I'll go with what Biswanath Rath said. That is an indefinite integral.

It is an indefinite integral, and "a logical answer" refers to its definition (what kind of I.I.?). If it converges, the l'Hospital rule may be applied.

Also some strange answer, probably you have to simplify the integrand first.

@Amir,

please follow the advice of Issam and give your formula explicitely!

I can't see how anyone can expect to get a useful answer to such a carelessly formulated question.

Recently I had this problem in my research. The conditions for L'Hospitals rule were satisfied. I had to repeat the rule two or three times to finally get a good result that was not indeterminate. You can get varied results so make sure you understand the nature of the functions before you go to the limit.

@ Amir,

Could you please respond to the follower's request:

Set up the exact form of the integral to solve the problem.

See all above but first, check your math very carefully, especially your initial mathematical model of what you are trying to analyze. 0/0-0/0 suggests a doubly indeterminate result. One possibility is an inadequate initial mathematical formulation of the situation being modelled.

This integral is improper integral and you must apply L'Hobital Rule. You send the integral for me

The cronies are right, you can not close your eyes to using L'Hospital's rule. There are examples where the use of this rule leads to the non-existence of a limit, in fact, there is a limit.

@Amir,

if you could manage to communicate your integral (I'm aware that humans can be in conditions, e.g. illness, that don't allow them to follow iterated requests from strangers) I would feed the expression to the program 'Mathematica' and let either you or the community know the result.

If the conditions of Lompital's lemma are not satisfied, then it can not be used. There are examples in which the limit exists, but the use of this rule leads to no existence of a limit.

@ Bahtiyar,

Can one apply some rule without satisfying its conditions?

I think you mean, the wrong use of L'hopital's rule leads to a contradiction.

Yes. More precisely, they do not verify the fulfillment of the conditions of the lemma.

To use L'Hopital's rule, you first combine the difference between ratios to obtain a single ratio (like combining fractions) of numerator divided by denominator. Depending on the example, this step alone might remove difficulties. But if the resulting ratio still has the property that numerator and denominator both approach zero, then use L'Hopital's rule.

If you could show us the function, there may be another way to do the integration rather than anti differentiation. Would you show us the function?

A further alternative method is to use some Taylor polynomial with a suitable remainder.

Again, the order requested to the polynomials, the methods are related to the nature of the problem.

Let your limits of integration be A and B, A the lower limit and B the upper limit (I will also assume AA from above and b->B from below.

I apologize for the delay to answer.

The problem is an integral of Fourier power series.

I have attached a picture of the equation.

The limits of integration is from 0 to b.

Dear Amir,

The complete solution of your integral is attached.

Just set the boundaries of the integral and evaluate directly.

There is no problem at all.

Best

Mathematica does not agree with Issam, as from Issam's third formula-line a factor 'b' dissapeared. As the appended pdf-file shows the result for each of the Integrals is simply 0.

I hope that all contributors who tried to give advice without knowing what the issue really is, will recognize the vanity of such an approach.

@ Ulrich,

Don't give a direct judge. :-)

It is the same answer as mine, doing manually.

Just read the steps where you can expand cos (A+B) and Cos(A-B) in my answer to obtain the same answer you obtained by Mathematica.

One needs to solve and write solutions for exact answers.

Mathematica and other packages are useful for long numerical calculations.

Elementary calculus can do the job.

P.S.

If one judge some work, he should make a little effort.

Don't trust machines all the time.

Best

@Issam,

sorry for having edited my answer during the creation of your's. So, as you can see, I meanwhile invested the little effort you asks for. You simply left out a factor b and thus got a wrong result. Of course, you are right by saying that Elementary calculus can do the job.

However, as the present example shows, it is not always easy to avoid lapses. Mathematica is not creative enough to make lapses. I found it also extremely useful in symbolic computations (such as in our example) - the older I get, the more so!

Dear Amir,

Thanks for being given the opportunity to answer your question, giving the condition of the task.

As I guessed in the answer one week ago, there is no uncertainty in your example (and no need in l'Hospital rule). Assuming that the integration limits are from 0 to b, all the terms should be divided into two groups:

n \neq q and n = q. The integrals of the first one members (as indicated by Ulrich Mutze) take zero values.

But for the second group, the average value of the function (sin^2 (z)) is non-zero, i.e. 1/2 times the length of the interval. Therefore, the answer is: b/2 \sum{c_{nn}}. Hope, this helps.

P.S. Mathematica is my friend, but my greatest friend is mathematics (no sarcasm).

@ Ulrich,

I read your answer generated by Mathematica.

Unfortunately, it is the wrong answer. Notice that in your answer,

b disappeared outside the entries of sin and cos !!

I think you made a mistake when you entered the symbols.

The exact and the right answer is exactly the one attached to my previous answer.

You can regenerate the answer with right entries and compare with mine, the simplifications in the attached file help you to compare.

@Issam,

your Simplification.pdf starts with an expression in which the omission has already happened!

In your original Amir__Integral.pdf, in the third formula-line, there is a term

cos((n-q) pi y/2)

which obviously should be

cos((n-q) pi y/(2 b))

and accordingly for the term containing (n+q).

Hope, you see the point.

@ Ulrich,

Exactly, this what I asked you to take into consideration.

The missing b in the third line ( Just typo)

BUT, the next line, where calculations are done manually, give the exact answer, that introduces b again in the right place and solves the integral.

Recheck it to see the point.

@Issam,

you have simply to do the computation right (without omitting factors and re-introducing them in an illegal manner) to see that Amir's expression is simply a sum of zeros and thus 0.

Hope the appended file will convince you.

@Issam,

you where right in your distrust in the Mathematica result. For n=q it is obviously wrong as also pointed out in Volodymyrs's contribution.

For n different from q it is still right.

Actually Amir's expression is

\sum{c_{nn}} (b/2)(1 - sin(2 n pi)/(2 n pi))

where the factor to c_{nn} can be red from the appended file

@ Ulrich,

Trust my calculations. All results are direct as an elementary integrals.

Nothing more to do.

Just set the boundaries of the integral according to Amir,s model.

@Issam,

now your calculation is correct. Actually it was so from the very beginning, but was oscured by an unhappy typo. Your comments on this made me think that you follow a queer logic. Actually I misunderstood you. Nevertheless, stating the final result for Amir's expression would have been adequate.

For me the discussion was beneficial since it reveiled a malfunction of Mathematica that will communicate to Wolfram Research.

Thank you for the discussion!

@ Ulrich,

I am happy to join professional mathematicians as you and others.

This help to share knowledge at RG platform.

Thank you, Ulrich, :-)

I have better things to do than to answer all that crap

@Followers,

I think RG should be serious platform devoted to sharing knowledge for all levels, undergraduate and graduate students, MS, Ph.D., and professors.

All questions for all levels should be respected.

Some members use words that describe his point of view; such words may reflect his thoughts and personality as well.

I got the impression that the question was a request for help. People that ask for help should be given help. Not insults. Although my attempt was not helpful, I am very happy that some other people provided the needed help. So RG does work, even if there are a few members that are less constructive.