C.11 · Integration by parts, order reversal and symmetry

Undergraduate GRE preparation. Local objectives within the reviewed ETS scope; this is original teaching, not an official test or score predictor.

Prerequisites: Product rule, improper limits and planar integration regions.

• 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

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.

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.

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.

To reverse ∫₀¹∫ₓ¹F(x,y)dy dx, describe the triangle 0≤x≤y≤1, then rewrite it as ∫₀¹integral from 0 to yF(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.

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.

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.

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.

Evaluate $\int_0^1\int_x^1 e^{y^2}\,dy\,dx$ by reversing order. Describe the region before and after reversal. Derive a formula for $I_2(x)=\int_1^x(\ln t)^2\,dt$, $x>0$.

Worked solution

The region is $0\le x\le y\le1$. In the reversed order, y ranges from zero to one and x from zero to y. Continuity on the compact triangle permits order reversal, giving $\int_0^1 ye^{y^2}\,dy=(e-1)/2$. Integration by parts with $u=(\ln t)^2$, $dv=dt$ gives $I_2=x(\ln x)^2-2I_1$. Since $I_1=x\ln x-x+1$,

Differentiation checks the integrand; substituting x=1 checks the constant.

For positive integer m, evaluate $\int_0^{\pi}\sin(2mx)/\sin x\,dx$. Justify endpoint integrability before invoking symmetry.

Check after attempting

The endpoint limits are 2m at zero and -2m at pi, by the first-order sine limit. Both singularities are removable, so the continuously extended integrand is integrable. Reflection $x\mapsto\pi-x$ preserves the denominator and reverses the numerator. The integral equals its negative and therefore is zero. Symmetric cancellation would not establish convergence for a genuinely divergent endpoint; the removable limits are essential.