1. ## hard double integral

Im stuck on this double integral, i cant see how it can be done!

Outer limit 0 - pi/2 inner limit x - pi/2

siny/y dydx

also this one

Outer limit 0 - 1 inner limit e^y - e

x/lnx dxdy

thanks

2. Hi

$
\int_0^{\frac{\pi}{2}} \int_x^{\frac{\pi}{2}} y\cdot \sin y \, \mathrm{d}x \mathrm{d}y$
and $\int_0^1 \int_{\exp y}^{\exp 1} x\cdot \ln x \, \mathrm{d}x\mathrm{d}y$

Are these the right integrals ?

In both cases, you can first compute the inner integral...

$\int_0^{\frac{\pi}{2}} \int_x^{\frac{\pi}{2}} y\cdot \sin y \, \mathrm{d}x\mathrm{d}y= \int_0^{\frac{\pi}{2}} \int_x^{\frac{\pi}{2}} y\cdot \sin y \, \mathrm{d}y\mathrm{d}x=\int_0^{\frac{\pi}{2}}\left[\int_x^{\frac{\pi}{2}} y\cdot \sin y \, \mathrm{d}y\right]\mathrm{d}x$

Can you take it from here ?

3. For the second one, reverse integration order.

4. thankyou for your responses, though i must not have not typed them out properly, the functions are actually quotients:

siny / y dydx

x / Lnx dxdy

5. Well anyway in both cases reverse integration order.

6. ill try that for the 2nd one, though ive done that for the first one

outer limit: 0 to pi/2

inner limit: 0 to y

then should i use substitution to solve the integral?

7. then should i use substitution to solve the integral?
No, there is no need to do this : reversing integration order brings simplifications.

$\int_0^{\frac{\pi}{2}}\int_x^{\frac{\pi}{2}}\frac{ \sin y}{y}\,\mathrm{d}x\mathrm{d}y=\int_0^{\frac{\pi}{2 }}\left[\int_0^y\frac{\sin y}{y}\,\mathrm{d}x\right]\mathrm{d}y=\int_0^{\frac{\pi}{2}}\left[\frac{\sin y}{y}\int_0^y\,\mathrm{d}x\right]\mathrm{d}y=\ldots$

8. thanks for that flying squirrel took straight after that, was good , i got an answer of -1, hope im right :S

9. Originally Posted by sterps
i got an answer of -1, hope im right :S
I don't agree, the answer is 1. How did you integrate $y\mapsto \sin y$ ?