5.3 Improper Integrals Involving Rational and Exponential Functions


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1 Section 5.3 Improper Integrals Involving Rational and Exponential Functions dθ +a cos θ =, < a <. a sin θ ( dθ = a a b a + b cos θ b ), dθ a + b cos θ =, <a. a a + b dθ a + b sin θ = a a + b, <a. < b <a dθ a cos θ + b sin θ + c = dθ a cos θ + b sin θ + c = c a b, a + b <c. (a + c)(b + c), < c < a, c < b. 5.3 Improper Integrals Involving Rational and Exponential Functions In this section we present a useful technique to evaluate improper integrals involving rational and exponential functions. Let a and b be arbitrary real numbers. Consider the integrals () b f(x) and a f(x), where in each case f is continuous in the interval of integration and at its finite endpoint. These are called improper integrals, because the interval of integration is infinite. The integral a f(x) is convergent if b lim b a f(x) exists as a finite number. Similarly, b f(x) is convergent if lim a a b f(x) exists as a finite number. Now let f(x) be continuous on the real line. The integral f(x) is also improper since the interval of integration is infinite, but here it is infinite in both the positive and negative direction. Such an integral is said to be convergent if both f(x) and f(x) are convergent. In this case, we set (Figure )
2 3 Chapter 5 Residue Theory Figure Splitting an improper integral over the line. () f(x) = lim a a f(x) + lim b b f(x). We define the Cauchy principal value of the integral (Figure ) a (3) P.V. f(x) = lim f(x), if the limit exists. a a f(x) to be The Cauchy principal value of an integral may exist even though the integral itself is not convergent. For example, a a x = for all a, which implies that P.V. x =, but the integral itself is clearly not convergent since x =. However, whenever f(x) is convergent, then P.V. f(x) exists, and the two integrals will be the same. This is because lim a a f(x) and lim a a f(x) both exist, and so Figure The Cauchy principal value of the integral. P.V. f(x) = lim a = lim a a ( a = lim = a a f(x) f(x) + a f(x). a f(x) + lim a ) f(x) a f(x) Because of this fact, we can compute a convergent integral over the real line by computing its principal value, which can often be obtained by use of complex methods and the residue theorem. The following test of convergence for improper integrals is similar to tests that we have proved for infinite series. We omit the proof.
3 Section 5.3 Improper Integrals Involving Rational and Exponential Functions 3 PROPOSITION INEQUALITIES FOR IMPROPER INTEGRALS Let B A f(x) represent an improper integral as in (), where A = or B =. (i) If B A f(x) is convergent, then B A f(x) is convergent and we have B A f(x) B A f(x). (ii) If f(x) g(x) for all A < x < B and B A g(x) is convergent, then B A f(x) is convergent and we have B A f(x) B A g(x). You are encouraged to use Proposition or any other test from calculus (such as the limit comparison test) to show that an improper integral is convergent. Once the convergence is determined, we may compute the integral via its principal value, as we will illustrate in the examples. EXAMPLE Evaluate Improper integrals and residues: the main ideas I = x x 4 +. Solution To highlight the main ideas, we present the solution in basic steps. Step : Show that the improper integral is convergent. Because the integrand is continuous on the real line, it is enough to show that the integral outside a x finite interval, say [, ], is convergent. For x, we have x 4 + x x = 4 x, and since x is convergent, it follows by Proposition that x x 4 + and x x 4 + are convergent. Thus x is convergent, and so x 4 + x R x (4) x 4 = lim + R x 4 +. Step : Set up the contour integral. We will replac by z and consider the function f(z) = z z 4 +. The general guideline is to integrate this function over a contour that consists partly of the interval [, R], so as to recapture the integral R x x 4 + as part of the contour integral on γ R. Choosing the appropriate contour is not x obvious in general. For a rational function, such as, where the denominator x 4 + does not vanish on thaxis and the degree of the denominator is two more than the degree of the numerator, a closed semicircle γ R as in Figure 3 will work. Since γ R consists of the interval [, R] followed by the semicircle σ R, using the additive property of path integrals (Proposition (iii), Section 3.), we write (5) I γr = f(z) dz = f(z) dz + f(z) dz = I R + I σr. γ R [, R] σ R Figure 3 The path and poles for the contour integral in Example. For z = x in [, R], we have f(z) =f(x) = R x x 4 + x x 4 + and dz =, and so I R =, which according to (4) converges to the desired integral as R. So, in order to compute the desired integral, we must get a handle on the other quantities, I γr and I σr, in (5). Our strategy is as follows. In Step 3, we compute I γr by the residue theorem; and in Step 4, we show that lim R I σr =. This
4 3 Chapter 5 Residue Theory will give us the necessary ingredients to complete the solution in Step 5. Step 3: Compute I γr by the residue theorem. For R>, the function f(z) = z z 4 + has two poles inside γ R. These are the roots of z 4 + = in the upper halfplane. To solve z 4 =, we write =e i ; then using the formula for the nth roots from Section.3, we find the four roots z = +i, z = +i, z 3 = i, z 4 = i (see Figure 3). Since these are simple roots, we have simple poles at z and z and the residues there are (Proposition (ii), Section 5.) Res (z )= and similarly z d = z dz (z4 + ) 4z 3 = z=z 4z = z=z z=z Res (z )= 4z So, by the residue theorem, for all R> (6) I γr = γr z z=z = i 4. 4( + i) = i 4, z 4 + dz =i( Res (z ) + Res (z ) ) =i i 4 =. Step 4: Show that lim R I σr =. For z on σ R, we have z = R, and so z z 4 + R R 4 = M R. Appealing to the integral inequality (Theorem, Section 3.), we have I σr = z 4 + dz l(σ R)M R = R σr z R R 4 = R /R 3, Step 5: Compute the improper integral. Using (5) and (6), we obtain = I R + I σr. as R. Let R, then I R x x 4 + and I σ R, and so x x 4 + =. Figure 4 An alternative path for the integral in Example. In Example, we could have used the contour γ R in the lower halfplane in Figure 4. In this case, it is easiest to take the orientation of γ R to be negative in order to coincide with the orientation of the interval [, R]. The five steps that we used in the solution of Example can be used to prove a the following geberal result.
5 Section 5.3 Improper Integrals Involving Rational and Exponential Functions 33 PROPOSITION INTEGRALS OF RATIONAL FUNCTIONS Let f(x) = p(x) q(x) be a rational function with degree q(x) + degree p(x), and let σ R denote an arc of the circle centered at with radius R>. Then (7) lim p(z) R σ R q(z) dz =. Moreover, if q(x) has no real roots and z, z,..., z N denote all the poles in the upper halfplane, then of p(z) q(z) (8) p(x) N q(x) =i j= Res ( p(z) q(z),z ) j. Here is a straightforward application. Figure 5 The path and poles for the contour integral in Example. EXAMPLE Evaluate Improper integral of a rational function (x + )(x + 4). Solution The integrand satisfies the two conditions of Proposition : degree q(x) = 4 + degree p(x) =, and the denominator q(x) = (x + )(x + 4) has no real roots. So we may apply (8). We have (z + )(z + 4) = (z + i)(z i)(z +i)(z i), and hence has simple poles at i and i in the upper halfplane (Figure 5). The residues there (z +)(z +4) are and Res (i) = lim(z i) z i (z i)(z + i)(z + 4) = (i)( + 4) = i 6 According to (8), Res (i) = lim (z i) z i (z + )(z i)(z +i) = i. (x + )(x + 4) =i( i 6 + i ) = 6. Integrals Involving Exponential Functions The success of the method of contour integration depends crucially on the choice of contours. In the previous examples, we used expanding semicircular contours. To evaluate integrals involving exponential functions, we will use rectangular contours. We will consider integrals of the form e ax e bx, where a < b, and c>. + c
6 34 Chapter 5 Residue Theory Since c> and e ax >, the denominator does not equal zero. Also the condition a < b guarantees that the integral is convergent (Exercise 8). EXAMPLE 3 Integral involving exponential functions Let be a real number >. Establish the identity (9) e x + = sin. Solution Step : As noted before, since >, the integral converges. The integrand leads us to consider the function f(z) = ez e z +, whose poles are the roots of e z + =. Since the exponential function is iperiodic, then e z = =e i z = i +ki z k = (k +) i, k =, ±, ±,.... Thus f(z) = (Figure 6). ez e z + has infinitely many poles at z k, all lying on the imaginary axis Step : Selecting the contour of integration. Our contour should expand in the xdirection in order to cover the entiraxis. To avoid including infinitely many poles on the yaxis, we will not expand the contour in the upper halfplane, as we did with the semicircles in the previous examples. Instead, we will use a rectangular contour γ R consisting of the paths γ, γ, γ 3, and γ 4, as in Figure 6, and let I j denote the path integral of f(z) over γ j (j =,..., 4). As R, γ R will expand in the horizontal direction, but the length of the vertical sides remains fixed at. To understand the reason for our choice of the vertical length, let us consider I and I 3. On γ, we have z = x, dz =, Figure 6 The path and poles for the contour integral in Example 3. () I = R e x +, and so I converges to the desired integral I as R. On γ 3, we have z = x+i, dz =, and using the iperiodicity of the exponential function, we get () I 3 = R +i e (x+i ) + = e i R = e e x i I. + This last equality explains the choice of the vertical sides: They are chosen so that the integral on the returning horizontal side γ 3 is equal to a constant multiple of the integral on γ. From here the solution is straightforward. Step 3: Applying the residue theorem. From Step, we have only one pole of f(z) inside γ R at z = i. Since ez + has a simple root, this is a simple pole. Using Proposition (ii), Section 5., we find and so by the residue theorem () I + I + I 3 + I 4 = Res ( e z e z +, i) = e i e i = e i i = e ei, γr e z ( e z e z dz =i Res + e z +, i) = i e i.
7 Section 5.3 Improper Integrals Involving Rational and Exponential Functions 35 Step 4: Show that the integrals on the vertical sides tend to as R. For I, z = R + iy ( y ) on γ, hence e z = e R e iy = e R, e z + e z =e R e z + e R, and so for z on γ f(z) = e z e z + e R+iy e R = er e R = e ( )R e. Consequently, by the MLinequality for path integrals, I = e γ z e z + dz l(γ ) e ( )R e =, as R. e ( )R e The proof that I 4 as R is done similarly; we omit the details. Step 5: Compute the desired integral (9). Using (), (), and (), we find that I ( e i) + I + I 4 = i e i. Letting R and using the result of Step 4, we get and after simplifying which implies (9). ( e i) e x + = e x + i = ie, i ( e i e i) = sin, There are interesting integrals of rational functions that are not computable using semicircular contours as in Example, such as (3) x 3 +. This integral can be reduced to an integral invovling expoential functions, by using the substitution x = e t. We outline this useful technique in the following example. EXAMPLE 4 The substitution x = e t Establish the identity (4) x + = sin, any real number >. Solution Step : Show that the integral converges. The integrand is continuous, so it is enough to show that the integral converges on [, ). We have x +, x
8 36 Chapter 5 Residue Theory Figure 7 Splitting an improper integral. and the integral is convergent since x <. Step : Apply the substitution x = e t. Let x = e t, = e t dt, and note that as x varies from to, t varies from to, and so I = x + = e t e t + dt = e x +, where, for convenience, in the last integral we have used x as a variable of integration instead of t. Identity (4) follows now from Example 3. The tricky part in Example 3 is choosing the contour. Let us clarify this part with one more example. We will compute the integral ln x. The x 4 + integral is improper because the interval is infinite and because the integrand tends to as x. To define the convergence of such integrals, we follow the general procedure of taking all onesided limits one at a time. Thus (Figure 7) ln x x 4 + = lim ɛ ɛ ln x b x 4 + lim + b ln x x 4 +. It is not difficult to show that both limits exist and hence that the integral converges. We will use the substitution x = e t to solve the problem. If you like, you could instead check the convergence after changing variables. EXAMPLE 5 Derive (5) Solution An integral involving ln x ln x = x Let x = e t, ln x = t, = e t dt. This transforms the integral into t e 4t + et dt = x e 4x +, Figure 8 The path and poles for the contour integral in Example 4. where we have renamed the variabl, just for convenience. In evaluating this integral, we will integrate f(z) = zez e 4z + over the rectangular contour γ R in Figure 8, and let I j denote the integral of f(z) on γ j. Here again, we chose the vertical sides of γ R so that on the returning path γ 3 the denominator equals to e 4x +. As we will see momentarily, this will enable us to relate I 3 to I. Let us now compute I γr = γ R f(z) dz. As you can check, f(z) has one (simple) pole at z = i 4 inside γ R. By Proposition (ii), Section 5., the residue there is So by the residue theorem ( ) i( + i) (6) I γr =i 6 Res ( ze z e 4z +,i ) i = 4 ei 4 + i) = i( 4 4ei 6. = ( + i) 8 = I + I + I 3 + I 4.
9 Section 5.3 Improper Integrals Involving Rational and Exponential Functions 37 Moving to each I j (j =,..., 4), we have I = γ ze z e 4z + dz = R x e 4x +. On γ 3, z = x + i, dz =, so using e i = i, we get I 3 = = i γ3 ze z e 4z + dz = R = ii + R x e 4x + + R (x + i )ex e i R e 4x +. e 4x + e 4x + To show that I and I 4 tend to as R, we proceed as in Step 5 of Example 3. For z = R + iy ( y ) on γ, we have z R + y R +, and so, as in Example 3, f(z) = ze z e 4z + = z e z e 4z + (R + ) e R e 4R = R + e 3R e. Consequently, by the MLinequality for path integrals I = ze γ z e 4z + dz l(γ )(R + ) e 3R e = (R + ), as R. e 3R e The proof that I 4 as R is done similarly; we omit the details. Substituting our finding into (6) and taking the limit as R, we get ( + i) 8 ( = lim I + I ii + R R x = ( i) e 4x + + e 4x + + I 4 e 4x +. ) Taking imaginary parts of both sides, we obtain our answer x e 4x + = 8. If we take real parts of both sides we get the value of the integral in (4) that corresponds to = 4. The substitution x = e t is also useful even when we do not use complex methods to evaluate the integral in t. For example, ln x x + = te t e t dt = + since the integral is convergent and the integrand function of t. Exercises 5.3 tet e t + = t e t +e t is an odd
10 38 Chapter 5 Residue Theory In Exercises 8, evaluate the given improper integral by the method of Example x 4 + =. (x + )(x 4 + ) = (x + ) 3 = (4x + )(x i) = 3 i 8. x 4 + x + = 3 (x i)(x +3i) = (x 4 + ) = 3 4 (x + i)(x 3i) = In Exercises 9, evaluate the given improper integral by the method of Example e 3x + = 3 3 x = = + e x e ax e bx + = 3 b sin a b ( < a < b) In Exercises 3, evaluate the given improper integral by the method of Example 5. In some cases, the integral follows from more general formulas that we derived earlier. You may use these formulas to verify your answer x 3 + = 3 3 x x 5 + = x x 3 + = 3 x 3 = sin ( 5 4. ) ln(x) x +4 = ln. ( ) ln x 3 x = Use the contour γ R in Figure 9 to evaluate x 3 +. (n)! = ( + x ) n+ n (n!) x x + = sin ( ) ( > ) x = ( < < ) (x + ) sin x ln(x) x 3 + = 7 + ln 3 3 ln(ax) x + b = ln(ab) (a, b > ) b (ln x) 3 x 3 = Figure 9 The path and poles for the contour integral in Exercise. [Hint: I γr = I + I + I 3 ; I 3 = e i 3 I ; and I as R.] 6. Follow the steps in Example to prove Proposition. 7. A property of the gamma function. (a) Show that for < <, x +x = sin.
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