Understanding and Interpreting Improper Integrals

Understanding and Interpreting Improper Integrals

by Souad Ayadi -
Number of replies: 11

 

  1. What is the main difference between a proper and an improper integral?

  2. Why do we need limits when defining an improper integral?

  3. Can you give a physical example where an improper integral naturally appears (e.g., in electrostatics or thermodynamics)?

  4. Do all improper integrals with infinity as a bound diverge? Explain your reasoning.

  5. Optional: Share one example of an improper integral you have encountered and explain whether it converges or diverges.

In reply to Souad Ayadi

Understanding and Interpreting Improper Integrals

by SMAIL ISLAM -
1. Proper integrals are finite and well-behaved; improper integrals involve infinity or blow-ups.


2. We use limits because the integral cannot be evaluated directly at infinity or at singular points.


3. Example: calculating the field of an infinite line charge in electrostatics.


4. No — some improper integrals converge if the function decreases fast enough; others diverge.
In reply to Souad Ayadi

Understanding and Interpreting Improper Integrals

by Mounir BOUMAD -
Prosper intégral are finit and thé fonction IS continuons
impropre intégral are Infinite
2-we use limite because WE cannot evaluate a fonction directly at Infinite
3-
4-no
5-intgral e^-x DX
In reply to Souad Ayadi

Understanding and Interpreting Improper Integrals

by KAWTHER KLALDI -
1. Main Difference A proper integral is calculated over a bounded interval with a bounded function, whereas an improper integral deals with either unbounded intervals (such as infinity) or unbounded functions.
2. The integral cannot be directly calculated at a point of discontinuity or at infinity, so we use a limit to study its behavior.
3. It appears in physics when calculating the electric potential of a very long (infinite) charged wire.
4. No some converge if the function decreases rapidly enough (such as the integral of 1/x²).
5. Example: The integral of 1/√x from 0 to 1 is an improper integral (a singularity at zero) 
In reply to Souad Ayadi

Understanding and Interpreting Improper Integrals

by MERKIDENE MAROUA -

1:Proper integral:

The integrand is finite and continuous on a finite interval

Improper integral:

At least one of the following occurs:

The interval is infinite (e.g. 

The function is unbounded at some point in the interval

These integrals are defined using limits.

2:Because the integral cannot be evaluated directly:

The upper or lower bound may be infinite

Or the function may blow up at some point

Limits allow us to:

Replace the problematic bound or point

3:Electrostatics:

The electric potential U due to a point charge is computed by integrating the electric field from a point to infinity

Thermodynamics:

The partition function often involves integrals over all energies

4:No.

Some converge, others diverge, depending on how fast the function decreases.

Examples:

Convergent:

Divergent:

So infinity alone does not imply divergence.

5: 

In reply to MERKIDENE MAROUA

Understanding and Interpreting Improper Integrals

by KAWTHER KLALDI -
السلام عليكم ورحمة الله بركاته اختي في الإجابة رقم 4 لم  تظهر الامثله ربما حدث خطأ ما 
In reply to KAWTHER KLALDI

Understanding and Interpreting Improper Integrals

by MERKIDENE MAROUA -

اجل 

لقد حاولة الاجابة على شكل ملف لكن الحجم كان اكبر 

🥰شكرا على الملاحظة 

الاجابة هي :

∫(from 1 to ∞) (1/𝑥²) 𝑑𝑥 converges,

while ∫(from 1 to ∞) (1/𝑥) 𝑑𝑥 diverges.

In reply to MERKIDENE MAROUA

Understanding and Interpreting Improper Integrals

by KAWTHER KLALDI -
Oki
 
هل يمكن ان تعطيني مثال اخر على التكامل الغير صحيح؟
السوال 5
In reply to KAWTHER KLALDI

Understanding and Interpreting Improper Integrals

by MERKIDENE MAROUA -
وبخصوص سؤال رقم 5 الاجابة كالآتي :
Example: ∫(from 0 to ∞) e^{-x} \, dx
· Reason it is improper: The interval is unbounded ( [0, \infty) ).
· Convergence check: Using the limit:
\lim_{b \to \infty} \int_{0}^{b} e^{-x} \, dx = \lim_{b \to \infty} (1 - e^{-b}) = 1
· Result: The integral converges and its value is 1.
In reply to Souad Ayadi

Understanding and Interpreting Improper Integrals

by NIHAD HENNICHI -
1/Proper Integral: Has finite limits and a bounded (continuous) function.
​Improper Integral: Has at least one infinite limit (\infty or -\infty) or a function that becomes infinite (vertical asymptote) within the interval.
2/We use limits because we cannot evaluate a function at infinity or at a point where it is undefined.
In reply to NIHAD HENNICHI

Understanding and Interpreting Improper Integrals

by NIHAD HENNICHI -
3/Calculating the energy required to move an object from a planet's surface to infinity involves an improper integral.
4/No. An improper integral with an infinite bound converges if the function approaches zero fast enough.
In reply to Souad Ayadi

Understanding and Interpreting Improper Integrals

by MAROUA AHMED ABDELMALEK -

1. Proper Integrals

A proper integral is defined when the limits of integration are finite and the function remains continuous throughout the interval.

2. Improper Integrals

An integral is called improper if at least one of the following conditions is satisfied:

One or both integration limits are infinite.

The integrand is undefined or becomes unbounded at some point within the interval.

3. Definition Using Limits

Improper integrals are evaluated by expressing them as limits.

This approach allows us to determine whether the integral converges or diverges.

4. Physical Applications

Improper integrals frequently appear in physics.

In electrostatics, they are used to compute electric fields generated by charge distributions extending to infinity.

In thermodynamics, they help calculate the total energy of systems occupying unbounded regions of space.

5. Convergence and Divergence of Improper Integrals

An improper integral with infinite limits does not always diverge.

Its behavior depends on how the integrand behaves as � becomes large.

If the function decreases sufficiently fast, the integral converges.

If the function decreases too slowly or remains large, the integral diverges.