Solutions Serie 1 - preliminary exercises

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1 D-MAVT D-MATL Prof. A. Iozzi ETH Zürich Analysis III Autumn 08 Solutions Serie - preliminary exercises. Compute the following primitive integrals using partial integration. a) cos(x) cos(x) dx cos(x) cos(x) dx = sin(x) cos(x) + sin(x) sin(x) dx = sin(x) cos(x) cos(x) sin(x) + 4 cos(x) cos(x) dx From this we get 3 cos(x) cos(x) dx = cos(x) sin(x) sin(x) cos(x) + c, c R = cos(x) cos(x) dx = 3 cos(x) sin(x) 3 sin(x) cos(x) + c, c R, b) where c (and c = c/3) is a constant of integration. sin (x) dx sin (x) dx = sin(x) sin(x) dx = sin(x) cos(x) + cos (x) dx ( = sin(x) cos(x) + sin (x) ) dx = sin(x) cos(x) + x sin (x) dx As before, we isolate the term we want to compute and we get sin (x) dx = x sin(x) cos(x) + c, c R = sin (x) dx = (x sin(x) cos(x)) + c, c R. Please turn!

2 c) sin(nx) sin(mx) dx, n m, n, m 0 sin(nx) sin(mx) dx = n cos(nx) sin(mx) + m cos(nx) cos(mx) dx n = n cos(nx) sin(mx) + m ( n n sin(nx) cos(mx) + m ) sin(nx) sin(mx) dx n = n cos(nx) sin(mx) + m m sin(nx) cos(mx) + n n sin(nx) sin(mx) Therefore ) ( m n sin(nx) sin(mx) dx = m n sin(nx) cos(mx) cos(nx) sin(mx) + c, n c R Now because n m the leading term is not 0, and we are allowed to divide by it and obtain m sin(nx) cos(mx) n cos(nx) sin(mx) sin(nx) sin(mx) dx = n m + c, c R.. Find the primitive functions of the followings. a) x + 3 x(x ) We can find an alternative expression by partial fraction decomposition. The denominator contains the linear factor x with multiplicity and the linear factor (x ) with multiplicity. Therefore we can find an expression of the form x + 3 x(x ) = A x + B x + C (x ) x + 3 = A(x ) + Bx(x ) + Cx = (A + B)x + ( A B + C)x + A Comparing the coefficients we get (degree 0) A = 3 (degree ) A + B = B = (degree ) A B + C = 0 C = 5 and thus x ( x(x ) dx = x ) x + 5 (x ) dx = 3 ln x ln x 5 + c, c R. x Look at the next page!

3 b) x5 + x In this case the numerator has degree higher than the denominator. Therefore we have to first use polynomial division finding x 5 + : x x 3 + x+ ( )(x 5 x 3 ) = x 3 + : x ( )x 3 x = x + x 5 + x = x3 + x + x + x. Now we can apply partial fraction decomposition to the last term to get x + x = x + (x + )(x ) = A x + + B x x + = A(x ) + B(x + ) = (A + B)x + (B A). Comparing coefficients as before yields A =, B = 3 so that x 5 ( ) + x dx = x 3 + x (x + ) + 3 dx (x ) = x4 4 + x ln x ln x + c, c R. c) + x 3 We can observe that for each n N + x n+ = ( + x) n ( ) k x k. k=0 In our case n = and + x 3 = ( + x)( x + x ) look, for example, on Wikipedia the page Polynomial long division. 3 Please turn!

4 leads to the Ansatz + x 3 = ( + x)( x + x ) = A + x + Bx + C ( x + x ) = A( x + x ) + (Bx + C)( + x) = (A + B)x + (B A + C)x + A + C. By comparing the coefficients (degree ) A + B = 0 B = A (degree ) B A + C = 0 C = A (degree 0) A + C = A = 3, B = 3, C = 3 and + x 3 dx = 3 + x dx + 3 = 3 ln( + x ) + x 3 x + x dx + c = 3 ln( + x ) x 6 x + x dx + x x + x dx x + x dx + c = 3 ln( + x ) 6 ln( x + x ) + 3 arctan ( x 3 ) + c, c R. 3. Solve the following linear differential equations: a) y (t) + y (t) 6y(t) = 0 Solving the characteristic polynomial provides the general solution b) 4y (t) + y (t) + 9y(t) = 0 λ + λ 6 = 0 λ =, λ = 3 y(t) = C e t + C e 3t, C, C C. As before we solve the characteristic polynomial and we get the general solution 4λ + λ + 9 = (λ + 3) = 0 λ, = 3 y(t) = C e 3 t + C te 3 t, C, C C. 4 Look at the next page!

5 c) y (t) 6y (t) + 3y(t) = 0 Again λ 6λ + 3 = 0 λ = 3 + i, λ = 3 i and general solution y(t) = e 3t (C e it + C e it ) = = e 3t ((C + C ) cos(t) + i(c C ) sin(t)) = = e 3t (C cos(t) + C sin(t)), C, C C, where C = C + C and C = i(c C ) are still arbitrary constants. 4. a) With the help of trigonometric addition formulas, verify that the following identity is true cos(αx) cos(βx) = (cos((α + β)x) + cos((α β)x)), for α, β, x R b) Find similar expressions for sin(αx) sin(βx) and sin(αx) cos(βx). The trigonometric addition formulas are sin(αx + βx) = sin(αx) cos(βx) + sin(βx) cos(αx) () sin(αx βx) = sin(αx) cos(βx) sin(βx) cos(αx) () cos(αx + βx) = cos(αx) cos(βx) sin(αx) sin(βx) (3) cos(αx βx) = cos(αx) cos(βx) + sin(αx) sin(βx). (4) Addition of (3) and (4) solves a). Subtraction of (3) from (4) produces a similar expression for sin(αx) sin(βx) = (cos((α β)x) cos((α + β)x)) and finally addition of () and () yields sin(αx) cos(βx) = (sin((α + β)x) + sin((α β)x)). 5 Please turn!

6 c) Prove for n, m N the orthogonality relation π 0, if n m cos(nx) cos(mx) dx = π, if n = m and n, m 0 π, if n = m = 0. In the case n m we use the identity from a). In this case both n m and n + m are different from 0, therefore we can proceed as follows π cos(nx) cos(mx) dx = π cos((n + m)x) dx + π cos((n m)x) dx = sin((n + m)x) π sin((n m)x (n + m) + π (n m) = = 0. In the case n = m 0 we also use the identity in a), but this time the second summand simplifies differently π cos(nx) cos(mx) dx = π cos((n + m)x) dx + π cos((n m)x) dx = π cos(mx) dx + π dx = 4m sin(mx) π + x π = 0 + π = π. The last one is obvious. 5. (Bonus exercise) a) Find a primitive of the secant function: sec(x) dx, remember that sec(x) = / cos(x). Depends on which method we use, but the possible final expressions we get, obviously all equivalent, should be sec(x) = cos(x) = (i) ln + sin(x) sin(x) + c 6 Look at the next page!

7 (ii) (iii) (iv) (v) ln + tan(x/) tan(x/) + c ln sec(x) + tan(x) + c ln tan(π/4 + π/) + c tanh (sin(x)) + c Hint: expression (iii) is the one usually found on calculus textbooks. There s a trick to get to it by multiplying and dividing by an appropriate expression, but it seems rather artificial, so we propose a way to get to (i), using the techniques we used before (basically partial fraction decomposition). sec(x) = cos(x) cos(x) = cos (x) = cos(x) sin (x) = cos(x) ( + sin(x))( sin(x)) =... where we suggest to fill the dots finding an alternate expression to the function ( + u)( u) via partial fraction decomposition, and apply it to u = sin(x). b) Prove that 0 x 4 ( x) 4 + x = 7 π. it s useful to simply expand the numerator (we get a polynomial of degree 8) and then use polynomial division. At this point we will have an expression involving sums of functions of which we know explicitely the primitives, and we can explicitely compute them in and 0 to get the final result. c) The Gamma function is the function Γ(α) = + 0 x α e x dx, α > 0. First, one should note that the function is well defined, that is the integral actually converges for each value of α > 0. (i) Prove that for each α > 0 the following recursion formula holds: (ii) Compute explicitely Γ() =. Γ(α + ) = αγ(α) 7 Please turn!

8 (iii) Deduce from (i) and (ii) that for integer numbers n N: Γ(n + ) = n! Therefore the Gamma function is a possible extension of the factorial function to all positive real numbers. (i) Hint: it s enough to integrate by parts, using f = x α, and g = e x. (ii) This is a straightforward computation. (iii) Proof by induction, starting with n = 0. (Remember that, by definition, 0! = ). 8

Chapter 7: Techniques of Integration

Chapter 7: Techniques of Integration MATH 206-01: Calculus II Department of Mathematics University of Louisville last corrected September 14, 2013 1 / 43 Chapter 7: Techniques of Integration 7.1. Integration