integral of xtan(pix)

解答

\int x tan (π x) d x

i \frac{1}{2 π ^{2}} Li_{2} (- e^{2 iπ x}) + \frac{i x ^{2}}{2} - \frac{1}{π} x ln (1 + cos (2 π x) + i sin (2 π x)) + C

求解步骤

\int x tan (π x) d x

使用换元积分法

= \int \frac{u tan ( u )}{π ^{2}} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{π ^{2}} \cdot \int u tan (u) d u

这是非初等函数积分:

\int x tan (x) d x = i \frac{1}{2} Li_{2} (- e^{2 i x}) + i \frac{1}{2} x^{2} - x ln (1 + e^{2 i x})

= \frac{1}{π ^{2}} (i \frac{1}{2} Li_{2} (- e^{2 i u}) + i \frac{1}{2} u^{2} - u ln (1 + e^{2 i u}))

u = π x

代回

= \frac{1}{π ^{2}} (i \frac{1}{2} Li_{2} (- e^{2 iπ x}) + i \frac{1}{2} (π x)^{2} - π x ln (1 + e^{2 iπ x}))

化简

\frac{1}{π ^{2}} (i \frac{1}{2} Li_{2} (- e^{2 iπ x}) + i \frac{1}{2} (π x)^{2} - π x ln (1 + e^{2 iπ x})) : i \frac{1}{2 π ^{2}} Li_{2} (- e^{2 iπ x}) + \frac{i x ^{2}}{2} - \frac{1}{π} x ln (1 + cos (2 π x) + i sin (2 π x))

= i \frac{1}{2 π ^{2}} Li_{2} (- e^{2 iπ x}) + \frac{i x ^{2}}{2} - \frac{1}{π} x ln (1 + cos (2 π x) + i sin (2 π x))

解答补常数

= i \frac{1}{2 π ^{2}} Li_{2} (- e^{2 iπ x}) + \frac{i x ^{2}}{2} - \frac{1}{π} x ln (1 + cos (2 π x) + i sin (2 π x)) + C

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