lec4-3 - Differentiation of integrals and Leibniz rule •...

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Unformatted text preview: Differentiation of integrals and Leibniz rule • We are interested in solving problems of the type d dx Z v ( x ) u ( x ) f ( x , t ) dt • Notice in addition to the limits depending on x , the function f ( x , t ) also depends on x • First lets take the case where the limits are constants ( u ( x ) = b and v ( x ) = a ) • In most instances we can take the derivative inside the integral (as a partial derivative) d dx Z b a f ( x , t ) dt = Z b a ∂ f ( x , t ) ∂ x dt Patrick K. Schelling Introduction to Theoretical Methods Example • Find R ∞ t n e- kt 2 dt for n odd • First solve for n = 1, I = Z ∞ te- kt 2 dt • Use u = kt 2 and du = 2 ktdt , so I = Z ∞ te- kt 2 dt = 1 2 k Z ∞ e- u du = 1 2 k • Next notice dI dk = R ∞- t 3 e- kt 2 dt =- 1 2 k 2 • We can take successive derivatives with respect to k , so we find Z ∞ t 2 n +1 e- kt 2 dt = n ! 2 k n +1 Patrick K. Schelling Introduction to Theoretical Methods Another example... the even n case • The even n case can be done of R ∞ t n e- kt 2 dt • Take for I in this case (writing in terms of x instead of t ) I = Z ∞-∞ e- kx 2 dx • We can write for I 2 , I 2 = Z ∞-∞ Z ∞-∞ e- k ( x 2 + y 2 ) dxdy • Make a change of variables to polar coordinates, x = r cos θ , y = r sin θ , and x 2 + y 2 = r 2 • The dxdy in the integrand becomes rdrd θ (we will explore this more carefully in the next chapter) Patrick K. Schelling Introduction to Theoretical Methods Example: continued • Then in polar coordinates we solve the I 2 integral, I 2 = Z 2 π Z ∞ e- kr 2 rdrd θ = π k • So we find I = p π k • To find R ∞ x 2 e- kx 2 dx , we take- 1 2 dI dk = ( 1 2 )( 1 2 ) π 1 / 2 k 3 / 2 • For R ∞ x 4 e- kx 2 dx = 1 2 d 2 I dk 2 = ( 1 2 )( 1 2 )( 3 2 ) π...
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This note was uploaded on 07/30/2011 for the course PHZ 3113 taught by Professor Staff during the Spring '03 term at University of Central Florida.

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lec4-3 - Differentiation of integrals and Leibniz rule •...

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