integral of xtan(pix)

Solution

\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

Solution steps

\int x tan (π x) d x

Apply u-substitution

= \int \frac{u tan ( u )}{π ^{2}} d u

Take the constant out:

\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

Use the nonelementary integral:

\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}))

Substitute back

u = π x

= \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}))

Simplify

\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))

Add a constant to the solution

= 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