# Stuck on an integral

1. Mar 2, 2006

### BananaMan

i have a double integral to reverse the order of and then integrate, i have reversed the order fine, however i am VERY stuck on the integration of the function in the first integral

integrate sin(y)/(x+y) dx between 0 and y

any pointers greately appreciated

2. Mar 2, 2006

### Galileo

You're integrating wrt x, right? So sin(y) is a contant as far as the integration is concerned.

3. Mar 2, 2006

### BananaMan

so it would just be sin(y) * integral 1/(x+y)

therefore sin(y) * ln(x+y)

if so im going to feel like a fool :P

4. Mar 2, 2006

### BananaMan

if this is the case i have now integrated it and subbed for (y) (from the limits) to get sin(y)*ln(2y) but i must now integrate that function wrt y, which is even harder than the first function >.< and i am very stuck, maybe integration by parts here?

5. Mar 3, 2006

### benorin

Don't feel like a fool; the thing is so. Post the bounds so we can help...

6. Mar 3, 2006

### BananaMan

ok the original integral was between y and 0 hence getting sin(y)*ln(2y)

now the integral is that function ive just typed between pi/2 and 0 dy

im muchos stuck

7. Mar 3, 2006

### dextercioby

$$-\int_{0}^{\frac{\pi}{2}} \sin y \ \ln 2y \ dy =-\left[ \mbox{Ci}\left( \frac{1}{2}\pi \right) +\ln 2-\gamma\right]$$.

Daniel.

8. Mar 3, 2006

### Galileo

The bounds on the first integral are y and 0 right? So just enter the bounds correctly. Don't skip too many steps or you might miss a simplification:
$$\ln(x+y)|^{x=y}_{x=0}=\ln(2y)-\ln(y)=\ln(\frac{2y}{y})=\ln2$$

Which doesn't even depend on y, so the integral is easy as pie.

9. Mar 3, 2006

### BananaMan

thanks :)

i just forgot to seperate the variables and made life hard on myself