1. ## Improper integral

Evaluate $\int_0^1\sin(\ln x)\,dx$

2. Originally Posted by liyi
Evaluate $\int_0^1\sin(\ln x)\,dx$
You can write, $\int_0^1 \sin (\ln x) dx = \int_0^1 x\sin (\ln x) \cdot (\ln x)' dx$

Now let $t=\ln x$ and we get, $\int_{-\infty}^0 e^t \sin t ~dt$

This integral is doable.

3. You could try it this way:

$\int_{0}^{1}sin(ln(x))dx$

$\lim_{L\rightarrow{1}}\int_{0}^{L}sin(ln(x))dx$

One way is you can use Integration by Parts for this:

Let $u=sin(ln(x)), \;\ dv=dx, \;\ du=\frac{cos(ln(x))}{x}, \;\ v=x$

Then you get $xsin(ln(x))-\int{cos(ln(x))}dx$

Now, use parts again:

$u=cos(ln(x)), \;\ dv=dx, \;\ du=\frac{-sin(ln(x))}{x}, \;\ v=x$

Putting it together, we end up with:

$\lim_{L\rightarrow{1}}\left[\frac{Lsin(ln(L))}{2}-\frac{Lcos(ln(L))}{2}\right]$

Now, as you can see, as L approaches 1, we have:

$0-\frac{1}{2}=\frac{-1}{2}$

4. Originally Posted by liyi
Evaluate $\int_0^1\sin(\ln x)\,dx$
Double integration leads TPH's integral too.

$\sin (\ln x) = \int_0^{\ln x} {\cos u\,du} .$ So

$\int_0^1 {\sin (\ln x)\,dx} = - \int_0^1 {\int_{\ln x}^0 {\cos u\,du} \,dx} = - \int_{ - \infty }^0 {\underbrace {\int_0^{e^u } {dx} }_{e^u }\cos u\,du} .$

Hence $\int_0^1 {\sin (\ln x)\,dx} = - \int_{ - \infty }^0 {e^u \cos u\,du} .$

Now this is routine, substitute $u=-v,$

$- \int_0^\infty {e^{ - v} \cos v\,dv} = - \text{Re} \int_0^\infty {e^{ - (1 - i)v} \,dv} = - \text{Re}\, \frac{1}
{{1 - i}} = - \frac{1}
{2}.$

The rest follows.