Tích phân từng phần, đảo thứ tự và tính đối xứng

Change the order, change the difficulty

• $\displaystyle\int_0^1\!\!\int_x^1 e^{y^2}\,dy\,dx$ has no elementary inner antiderivative. Reverse the order and it becomes $\int_0^1 y e^{y^2}\,dy = \frac{e - 1}{2}$.
• GRE integration tests parts (with boundary terms 边界项), order reversal and symmetry 对称性.

Integration by parts 分部积分

• $\int u\,dv = uv - \int v\,du$; for definite integrals evaluate $uv$ at both ends.
• $I_n(x) = \int_1^x (\ln t)^n\,dt$: $u = (\ln t)^n$, $dv = dt$ gives $I_n = x(\ln x)^n - nI_{n-1}$. Hence $I_2 = x[(\ln x)^2 - 2\ln x + 2] - 2$.

Improper integrals by parts

• Apply parts on finite bounds, then take limits — the boundary term must vanish.
• With $\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt\pi$: $u = x$, $dv = xe^{-x^2}dx$ gives $\int_{-\infty}^{\infty} x^2e^{-x^2}dx = \frac{\sqrt\pi}{2}$, and by evenness $\int_0^\infty = \frac{\sqrt\pi}{4}$.

Reversing the order

• Draw the region first. $0 \le x \le y \le 1$ is $\int_0^1\int_x^1 \cdots\,dy\,dx = \int_0^1\int_0^y \cdots\,dx\,dy$.

Symmetry

• If $f(a + b - x) = -f(x)$ on $[a,b]$ and $f$ is integrable, $\int_a^b f = 0$.
• $\int_0^\pi \frac{\sin 2mx}{\sin x}dx = 0$: $x \mapsto \pi - x$ flips the numerator only. The endpoint singularities are removable (limits $2m$ and $-2m$) — check that first.

Recap