Evaluate the following improper integrals as iterated integrals.
(a) Z ∞
1
Z 1
e−x
x
3y
dy dx (b) Z 1
−1
Z √ 1
1−x2
√−1
(2y + 1) dy dx
a)
∫1∞∫e−x11x3ydydx=∫1∞∫e−111x3ydydx=∫1∞lnyx3∣e−11dx=\intop^{\infin}_1\intop^1_{e^{-x}}\frac{1}{x^3y}dydx=\intop^{\infin}_1\intop^1_{e^{-1}}\frac{1}{x^3y}dydx=\intop^{\infin}_1\frac{lny}{x^3}|^1_{e^{-1}}dx=∫1∞∫e−x1x3y1dydx=∫1∞∫e−11x3y1dydx=∫1∞x3lny∣e−11dx=
=∫1∞1x3dx=−12x2∣1∞=−1/2=\intop^{\infin}_1\frac{1}{x^3}dx=-\frac{1}{2x^2}|^{\infin}_1=-1/2=∫1∞x31dx=−2x21∣1∞=−1/2
b)
∫−11∫−1/1−x21/1−x2(2y+1)dydx=∫−11∫−∞∞(2y+1)dydx=∫−11(y2−y)∣−∞∞dx=−∞\intop^{1}_{-1}\intop^{1/\sqrt{1-x^2}}_{-1/\sqrt{1-x^2}}(2y+1)dydx=\intop^{1}_{-1}\intop^{\infin}_{-\infin}(2y+1)dydx=\intop^{1}_{-1}(y^2-y)|^{\infin}_{-\infin}dx=-\infin∫−11∫−1/1−x21/1−x2(2y+1)dydx=∫−11∫−∞∞(2y+1)dydx=∫−11(y2−y)∣−∞∞dx=−∞
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