Integration by parts, order reversal and symmetry
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| integration by parts/ˌɪntɪˈɡreɪʃn baɪ pɑːts/ | 分部积分 | fēn bù jī fēn |
| reflection symmetry/rɪˈflekʃn ˈsɪmətri/ | 反射对称 | fǎn shè duì chèn |
A decision before an answer
- An integral can become simple after changing its representation. A hard inner integral may turn into an elementary one when you reverse the region’s integration order.
- Your goal: Derive integration-by-parts reductions with valid endpoint limits.
Read the relationship
- Integration by parts comes from the product rule: ∫u dv=uv−∫v du. For definite integrals include the boundary term at both ends. With I_n(x)=∫₁ˣ(ln t)^n dt, choose u=(ln t)^n and dv=dt. For n≥1 the lower boundary vanishes, giving I_n=x(ln x)^n−n I_(n−1). A plus sign would violate the differentiated identity. Keep the same lower limit in a recurrence; changing it changes the constants.
- Reverse a double integral by reconstructing its region.
If I_n(x)=∫₁ˣ(ln t)^n dt for n≥1, which recurrence is correct?
Integration by parts gives uv minus the remaining integral. The lower boundary is zero for n≥1.
Use the defining rule
- For improper integrals first apply the identity on finite endpoints, then justify the limiting boundary and remaining integral. If ∫₋∞^∞e^(−x²)dx=√π, set u=x and dv=x e^(−x²)dx for ∫x²e^(−x²)dx. Since v=−e^(−x²)/2 and x e^(−x²)→0 at both infinities, the full second moment is √π/2. Evenness gives the positive-half-line moment √π/4. The vanished boundary is part of the argument, not an automatic property of every improper integral.
- Use reflection symmetry while checking removable endpoint behaviour.
After reversing ∫₀¹∫ₓ¹e^(y²)dy dx, what integral remains?
At each y, x ranges from 0 to y. The inner integral supplies length y.
Check the conditions
- To reverse ∫₀¹∫ₓ¹F(x,y)dy dx, describe the triangle 0≤x≤y≤1, then rewrite it as ∫₀¹∫₀ʸF(x,y)dx dy. Both the outer interval and inner bounds change; swapping symbols alone changes the region. For F=e^(y²), integrating over x first gives ∫₀¹y e^(y²)dy=(e−1)/2. Continuity on this compact triangle makes order reversal valid; singular or conditionally convergent cases require stronger care.
- Use reflection symmetry while checking removable endpoint behaviour.
For I₂(x)=∫₁ˣ(ln t)²dt, the recurrence gives x(ln x)²−2I₁(x), with I₁(x)=x ln x−x+1. Therefore I₂=x[(ln x)²−2ln x+2]−2. The constant −2 ensures I₂(1)=0. Separately, reversing the triangular e^(y²) integral produces a factor y from the inner x-length; dropping that factor would restore the original difficulty and change the value.
For I₂(x)=∫₁ˣ(ln t)²dt, the boundary value I₂(1) is ____.
The upper and lower integration endpoints coincide, and the explicit recurrence has the same zero boundary value.
Apply the task format
- If an integrable function on [a,b] satisfies f(a+b−x)=−f(x), reflection makes its integral equal to its negative, hence zero. For sin(2mx)/sin x on [0,π] with positive integer m, reflection x↦π−x changes the numerator sign and preserves the denominator. The apparent endpoint singularities are removable: limits are 2m at zero and −2m at π. Check these limits before invoking symmetry; cancellation is not a substitute for integrability.
- Use reflection symmetry while checking removable endpoint behaviour.
Improper integration by parts needs endpoint limits. Reconstruct the region before reversing order. An odd-looking reflected integrand must be integrable, including any removable endpoints.
Which answer fits this case?
Derive integration-by-parts reductions with valid endpoint limits
Integration by parts on a finite interval automatically remains valid after replacing its endpoints with infinities.
The boundary term and remaining integral must have the required limits. Divergent or oscillating boundaries invalidate an automatic replacement.
Keep the distinctions
- integration by parts 分部积分 — An integration identity derived from the product rule with a boundary term.
- reflection symmetry 反射对称 — A relation between integrand values at points mirrored across an interval midpoint.
- Derive integration-by-parts reductions with valid endpoint limits.
- Reverse a double integral by reconstructing its region.
- Use reflection symmetry while checking removable endpoint behaviour.
Match each term with its precise meaning in this lesson.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.