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By Otto Mutzbauer

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1 1 = 1 + 32 + 52 + .... J2rr 11 (11 le)n Let f(x) that j, = In x, where 1 < x < 11. 00 . x"- -{(In 1 + In2) + (ln2 + In 3) + .. ·+[In(11 - 1) + Inn}} 2 1 = "error terms. ) where En has a limit as 11 -+ 00. Conclude that n! fii(n/e)n . C --, , as 11 -+ = (n+~) Inn-l1 +E 00, Then by Wallis's formula, we have S; 2211 (111)2 -=h -+~. jiZ(211 ) ! J2n:n(nle)"] -+ 1 as 11 -+ 00. , r:; vn: e-x - dx = - , /, o 2 C > O. n II , 32 A Garden of Integrals Using integration by parts, show that 00 /, _ 2 xn+2e x dx = 11 a + 1 /,00 xne_x2 dx.

Now, suppose we have two windows, (al' b I] at tl and (a2. b2] at t2, where a < 11 < t2 < 1. See Figure 25. We have Wiener assigned a measure of If the window at time tl is large, (-co, co], no real restriction is imposed on the number of particles, and the measure of 24 A Garden of Integrals position x Figure 25. should be the same as the measure of {xC·) that E Co I a2 < X(t2) < b2}. Show Similarly if (a2, b21== (-00,00]. : 00, -00 < x (t2) < 00,0 < tl < t2 ~ In =1. Finally, let K(x t t) == (2nt)-1/2 e-X /2t, with 0 < t1 < t2 < ...

Sino(rc/ x) 0 < x < 1, x =0. a. Calculate F'. b. Show that F' is continuous on [0, 1]. c. Calculate C Jot F' (x) dx. Cauchy not only gave us the existence of the integral for a large class of functions (continuous), but also gave us a straightforward means of calculating many integrals. 4 Recovering Functions by Differentiation . In addition to the idea of recovering a function from its derivative by integration, we have the notion of recovering a function from its integral by differentiation, the second part of the Fundamental Theorem of Calculus.

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