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3 solutions
Yup! The way to calculate ∫ − ∞ ∞ f ( x ) is to take
a → − ∞ , b → ∞ lim ∫ a b f ( x ) d x .
The integral exists if and only if the limit exists.
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can i learn calulus
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Of course you can. Nothing's stopping you!
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@Calvin Lin – no i meant from u
But why did you choose 2n of all numbers? I see that 2 times infinity=infinity..
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He can choose any value of a , b (refer to my comment) as long as they both tend to infinity.
But if you take two different limits (-n and 2n) the answer will differ,, isn't that obvious?
indeterminate integral? :)
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Yes. This is an indeterminate. Any difference between the -infinity and the +infinity can make the result to change significantly.
However, it is customary to define the principal value integral as the first limit you present. The principal value of this integral is defined: it is equal to zero.
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I don't think it is customary. It is one version that is used to attempt to deal with such infinite integrals when they appear, in order to allow us to reasonably manipulate them further. Also, when they are used, it is often denoted as P V ∫ − ∞ ∞ , instead of just the ∫ − ∞ ∞ .
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That is true--I am used to the notation P ∫ − ∞ ∞ .
From the line on x y-plane, we can imagine to see a straight line with areas of -infinity and +infinity respectively UNDER the line with respect to +d x or +x. With [(1/2) x^2] from -infinity to +infinity, we have no idea of how much to how much. As both ways are not converged, the integral is NOT determine-able.
"Does not exist" seem to be cruel. I would rather take "Indeterminate" instead of this.
So, with reference to our previous conversation about "rearrangement of terms arbitrarily in a sequence and evaluating the sum", why can't we make the + x cancel out with − x to obtain 0? You are holding 2 contradicting opinions in these 2 problems.
Update: I just saw your recent comment on the report. Keep thinking about this and resolve your doubts.
The correct mathematical phrase is "Does not exist". It has nothing to do with "being cruel". Indeterminate forms are those whose limits, upon substitution of algebraic values, yield the form of 0 0 , 0 × ∞ , 0 0 , 1 ∞ , … .
∫
b
a
(
x
)
dx
=
∫
b
a
(
a
+
b
−
x
)
dx
Applying above property to the given integral,
∫
−
∞
∞
(
x
)
dx
=
∫
−
∞
∞
(
∞
−
∞
−
x
)
dx
As, we know
∞
−
∞
is indeterminate. .
Therefore,given integral doesn't exist.
Several issues with this solution:
- Is that identity true for a , b = ± ∞ ? I don't think so.
- In order to apply that identify, you are first assuming that the integral exists. So, you cannot use it to prove that it doesn't exist.
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I didn't understand your second point. We are taught in lower classes that \sqrt{2} is irrational and generally the proof is given by a method called proof by contradiction where we first assume that \sqrt{2} is rational and later in the proof some contradiction occurs and we say \sqrt{2} is irrational. So why are you saying in this case we cannot use contradiction?
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Sure, it's possible to prove by contradiction, but that is not what he did in his proof. I am only commenting with respect to what he wrote, since i am not a perfect mind-reader.
For example, a valid proof could be along the lines of:
1. Assume that the integral exists and is finite.
2. Some steps.
3. Hence, contradiction.
4. Thus, the integral does not exist.
Note that in "2. some steps", we have to be careful and apply the relevant identity, which is another point in my original comment.
For example, it is true that ∫ − ∞ ∞ 0 d x exists and is finite (and in fact 0). However, we cannot apply the identity to say that
∫ − ∞ ∞ 0 d x = ∫ − ∞ ∞ ( ∞ − ∞ − 0 ) d x
and hence conclude that the integral does not exist.
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@Calvin Lin – How is ∫ − ∞ ∞ 0 d x equal to zero?
I think the method of contradiction is different from proving existence. In your proof using contradiction, you assumed existence already.
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@Christ Ian Badana – Assuming that sqrt(2) is rational is different from assuming that "something" exists.
I think you are right sir.
We can see this question by graph too...
Integration means area under the curve, area under the curve for given integral is infinite positive as well as infinite negative which is not possible to calculate.Hence,integral doesn't exist.
Is this right?
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Not quite, but close. The interpretation of "area under the curve" is valid for Riemann integrals, but not necessarily for other types of integers.
The idea of "there is infinite positive as well as infinite negative area" is the crucial part here.
Hint: What do we really mean when we say ∫ − ∞ ∞ f ( x ) d x ? What limit are we taking?
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@Calvin Lin – Do you mean that given integral doesn't exist because we are integrating a function f ( x ) from those points which it can't achieve...
I think the integral exists but it is not defined. Once cannot say that there is no area bound by the curve under the given limits.
If your solution was correct then we would not have been able to calculate ∫ 0 ∞ sin ( x ) / x \dx , but we can.
To highlight why the answer does not exist, we can take a limiting process to get two different answers.
n → ∞ lim ∫ − n n x d x = n → ∞ lim [ 2 n 2 − 2 ( − n ) 2 ] = 0
n → ∞ lim ∫ − n 2 n x d x = n → ∞ lim [ 2 ( 2 n ) 2 − 2 ( − n ) 2 ] = n → ∞ lim 2 3 n 2 = ∞
Since it's not possible for the integral to have two different answers, the integral doesn't exist.