1873_solutions

# Jaj an n j0 we observe that n n a jaj n j0 1 nj 1 j0

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Unformatted text preview: of a Partial Improper Riemann Integral are satisfied and so if y is any real number we have Ý Þ0 gy 2x exp x 2 sin 2xydx. Therefore, for each y we have f y  2y exp y 2 g y  exp y 2 g y Þ0 Ý  2y exp y 2 g y exp y 2  2y exp y 2 g y exp y 2 wim lÝ 2x exp x 2 sin 2xydx Þ 0 2x exp w x 2 sin 2xydx We now apply the method of integration by parts to the last integral and obtain lÝ f y  2y exp y 2 g y  exp y 2 wim exp w 2 sin 2wy  2y exp y 2 g y exp y 2 Þ0 Ý Þ 0 2y exp w 0 2y exp x 2 cos 2xydx x 2 cos 2xydx 0 and we conclude that the function f is constant. To find the value of this constant we observe that Ý  f 0  exp 0 2 g 0  Þ exp x 2 dx  . 2 0 3. Find an explicit formula for the integral Þ0 Ý exp x 2 cos 2xydx. From Exercise 2 we see at once that if y is any real number then Ý  Þ 0 exp x 2 cos 2xydx  2 exp y 2 . 4. Evaluate the integral 384 Ý Þ0 Þ0 Ý exp x 2 cos 2xydxdy What happens in this integral if we invert the order of integration? Ý Ý Ý  Þ 0 Þ 0 exp x 2 cos 2xydxdy  Þ 0 2 exp y 2 dy   . 4 If we invert the order of integration, the inside integral diverges. Since the order of integration can’t be inverted we know that the given integral converges conditionally. 5. Apply Fichtenholz’s theorem to the integral Þ Ý 0 Ý Þ 0 f x, y dxdy where exp y 3 if x  y 2 0 f x, y  if x . y2 Now evaluate the integral Ý Þ0 Þ exp y 3 dydx. x We apply the version of the Fichtenholz theorem for improper integrals to obtain Ý Þ0 Þ0 Ý Þ0 Þ0 Ý Ý Ý f x, y dxdy  Ý f x, y dydx which gives us Ý y2 Þ0 Þ0 Þ0 Þ exp y 3 dxdy  exp y 3 dydx x and we conclude that Ý Þ0 Þ 6. Ý x exp y 3 dydx  Þ0 Ý e y 3 y 2 dy 1 3 a. Express the integrand of the following integral in partial fractions and show that if x and y are positive numbers then Ý Þ 0 1  t 2 x 2 1 1  t 2 y 2 dt  2 x y .  1 1 t2x2 1 t2y2  x2 y2 x2 1  t2x2 x2 y2 y2 1  t2y2 and so Þ0 Ý Þ0 x2  2 1 t2y2 dt  w Ý lim wÝ lim 1 1 t2x2 x2 w x2 x y2 y2 x2 dt 1  t2x2 arctan tx y x y2 x2 y2 lim wÝ  y x2 y2 x2 arctan ty  . 2 xy b. Apply Fichtenholz’s theorem (more than once) to the integral 1 1 Þ 0 Þ 0 2 x y dxdy  and deduce that 2 Ý Þ 0 arctan x dx   log 2. 2 x We begin by observing that 385 Þ0 w lim wÝ y2 y2 dt 1  t2y2 Þ 0 Þ0 1 1  dxdy  2 xy Þ 0 Þ0 Þ0 1 1 Ý Ý 1 dtdxdy 1  t2x2 1  t2y2  Þ 0 Þ0 Þ0 1 1 dxdtdy 1  t2x2 1  t2y2  Þ0 Þ 0 Þ0 1 dxdydt 1  t2x2 1  t2y2  Þ0 Þ  Þ0 1 Ý 1 Ý 1 Ý 1 0 1  t 2 y 2 1 arctan tdydt t arctan t t2 2 dt On the other hand, Þ 0 Þ0 1 1  dxdy   2 2 xy Þ 1 0 log 1  y log y dy   log 2. c. Evaluate the integrals /2 Þ0 x 2 dx sin 2 x /2 Þ0 and x cot xdx Þ and /2 log sin xdx. 0 Solution: From the change of variable theorem we observe that whenever 0uv  2 we have Þu v x 2 dx  sin 2 x Þ tan u tan v arctan t 2 arctan t dt. sin 2 arctan t From the fact that sin 2 arctan t  cos 2 arctan t tan 2 arctan t 2 t2   t2 sec 2 arctan t 1t we deduce that Þu...
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## This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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