# can anyone evaluate the double integral for me?? int^(e_1) int^(e_1) ln (xy) dy dx

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### 1 Answer

You need to identify the inner integral such that:

`int_1^e ln(xy)dy`

Since the inner integral is `int_1^e ln(xy)dy` , you need to treat x as a constant.

You should evaluate this integral using parts such that:

`int udv = uv - int vdu`

`u = ln(xy) =gt du = x/(xy) dy =gt du = 1/y dy`

`dv = dy =gt v = y `

`int_1^e ln(xy)dy = (yln(xy))|_1^e - int_1^e y*(1/y)dy`

`int_1^e ln(xy)dy = (yln(xy))|_1^e - y|_1^e`

`int_1^e ln(xy)dy = y(ln(xy) - 1)|_1^e`

`int_1^e ln(xy)dy = e(ln(ex) - 1) - ln x + 1`

`int_1^e ln(xy)dy = e(ln e + ln x- 1) - ln x + 1 `

`int_1^e ln(xy)dy = e(1 + ln x - 1) - ln x + 1`

`int_1^e ln(xy)dy = eln x - ln x + 1`

You need to evaluate the outer integral such that:

`int_1^e (int_1^e ln(xy)dy) dx = int_1^e (eln x - ln x + 1) dx`

`int_1^e (int_1^e ln(xy)dy) dx = int_1^e ln x(e-1) dx + int_1^e dx`

`int_1^e (int_1^e ln(xy)dy) dx = (e-1)int_1^e ln x dx + x|_1^e`

You should evaluate the following integral using integration by parts such that:

`int_1^e ln x dx = (x ln x)|_1^e - int_1^e x*(1/x)dx`

`u = ln x =gt du = 1/x dx`

`dv = dx =gt v =x`

`int_1^e ln x dx = (x ln x - x)|_1^e `

`int_1^e ln x dx = eln e - e - ln 1 + 1`

`int_1^e ln x dx = e - e - 0 +1`

`int_1^e ln x dx = 1`

Substituting 1 for `int_1^e ln x dx` yields:

int_1^e (int_1^e ln(xy)dy) dx = e - 1 + e - 1

`int_1^e (int_1^e ln(xy)dy) dx = 2(e-1)`

**Hence, evaluating the double integral yields `int_1^e (int_1^e ln(xy)dy) dx = 2(e-1).` **