Evaluating Double Integral: Reversing Order and Simplifying by Parts

[chaiyar]

Apologies for not being proficient enough in the use of Latex to write this problem properly

I hope it will suffice if I simply describe it:

It is the integration of f(x,y)=(sin(y))/(x+y) with respect to x between limits 0 and y

which I've found to give ( sin(y) ) ( ln(2y) )

This must then be integrated with respect to y, which is where I'm having problems

Integrating by parts, which is the only way I can think of to do it, letting u=ln(2y), dv=sin(y)dy, v=-cos(y) and du=dy/y, just leaves another integral: (cos(y))/y dy

Would I be right in thinking the result of that last integral cos(y) over y has something to do with Taylor's series? This seems overly complicated though and besides the point of the question which is to reverse the order of the integral (which I did) and evaluate it. I'm sure I must be going about it the wrong way, perhaps its not meant to be done by parts?

I realize this is long-winded but I would be extremely grateful for any help.

Thanks very much!

 

[fzero]

Apologies for not being proficient enough in the use of Latex to write this problem properly

I hope it will suffice if I simply describe it:

It is the integration of f(x,y)=(sin(y))/(x+y) with respect to x between limits 0 and y

which I've found to give ( sin(y) ) ( ln(2y) )

You should recheck your calculation, since

$$\int dx \frac{\sin y}{x+y} = \sin y \ln(x+y) + c.$$

If you integrate from $$x=0$$ to $$x=y$$, the logarithmic factor is just a number, leaving a simple integral.

 

[chaiyar]

Ahh! You're right! Thanks very much!