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Please calculate $$I=\int_0^1 dx \int_0^x dy \int_0^y \frac{\sin z}{(1-z)^2}dz$$

Any hints? Thank you!

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Apply Fubini's Theorem. –  jip May 21 '13 at 6:09
    
Is this any different from $\int_0^1 \int_0^x \int_0^y {\sin (z) \over (1-z^2)} dxdydz$ ? Or are you having trouble algebraically ? –  hyg17 May 21 '13 at 6:50
    
@sizz,@hyg17. Thank you very much! Now I realized that it could be transformed to $\int_0^1 \frac{sin(z)}{(1-z)^2}dz\int_z^1 dy\int_y^1 dx$, just rewrite the domain of the integral, and it could be calculated. But how does the Fubuni Theorem work? –  Jiangnan Yu May 21 '13 at 6:56

1 Answer 1

The integral $$I=\int_0^1 \frac{\sin(z)}{(1-z)^2}dz\int_z^1 dy\int_y^1 dx\\=\int_0^1 \frac{\sin(z)}{(1-z)^2}dz\int_z^1(1-y)dy\\= \int_0^1 \frac{1}{2}\sin(z)dz=\frac{1}{2}-\frac{1}{2}\cos(1)$$

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The sign is wrong. The last integral should be +1/2 sin(z). –  Apprentice Queue May 21 '13 at 7:23
    
@ApprenticeQueue Thank you. :) –  Jiangnan Yu Jun 4 '13 at 12:10

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