The tricks one uses to evaluate integrals are so numerous that we cannot hope to cover them all here. There are however two fundamental ploys that are frequently employed. These are
• Substitution or change of variable.
• Differentiating with respect to a parameter.
As an example of the former consider F(xt, x2) = 1:r2 fdx
:J!j
(2.2.1)
1x2 x2dx
- Xt (x3 + 4)2. (2.2.2)
Let us say we only know to integrate powers of x. The integrand f here does not have that form. We will now bring it to that form by a change of variable. Let
(2.2.3) Our original goal was to plot f as a function of x, chop the region between Xt and x 2 into segments of width dx, evaluate the products f ( x )dx over the segments and add them all up in the appropriate limit. Our hope is that the job will be simpler in terms ofu. Suppose we express fin terms ofu and plot f{u). Along the u-axis the points u1 = x~ and u2 = x~ define the limits. Every value that f took as a function of x it will now take as a function of u at the corresponding point u = x3.
We do not however want the area under f(u) between u1 and u2 .• i.e., we do not want J f(u)du. We want 2:::1 f(ui)dx, which is the old area we began with. Thus given two nearby points separated by du, we want to form the product f ( u )dx, where dx is the corresponding separation along the x-axis. It is clear that since x is a function of u
dx = -du. dx (2.2.4)
du
Thus the integrand for the u-integration is I ( u )dx I du, where the factor dx I du is to be expressed in terms of u. Going back to our problem
u = x3 (2.2.5)
dn = 3x2 (2.2.6)
dx dx = 3x1 2 = 3u2/3 (2.2.7)
du dn du
(2.2.8) f(x)dx -> f(x(u)) ã 3u2/3 = 3{u + 4)2
}.,r2
1 f(x)dx = (2.2.9)
Switching once more to v = u + 4 we finally obtain
{X2 X2dX rv2=x~+4 dv
}.,1 (x3 + 4)2 = lv 1 =x~+4 3v2 ã (2.2.10)
It is now trivial to integrate l/v2 between the given limits.
Let us recall the main point in the above manipulations. Given a difficult integrand, we hope that by going to a new variable, we can end up with the simpler integrand. The new integrand is however not just the old one expressed in terms of the new variable, but that times the Jacobian
J (~) = dx.
u du (2.2.11)
Thus
1X1 X2 f(x)dx = 1u(x2) u(xt) f(x(u))J (~) U du (2.2.12)
with the limits expressed in terms of the corresponding variables.
Our mission is accomplished only if the new integrand is simple. In our example, the denominator of the original integrand did simplify upon going to v = x3 + 4, but the numerator did not. But fortunately it got canceled by the Jacobian. Conversely, the substitution of v would not have been very effective without the x2 in the numerator.
Problem 2.2.1. Evaluate J:r:t rx2 ...,....$_ vaã-x ... by switching to 9 defined by x = a sin 9.
Assume 0 < x < 1r /2.
Problem 2.2.2. Show J000 .,2~02 = ;a by switching to 9 defined by x = a tan 9.
Make sure the change of variables is sensible, namely, that every x in the range of integration be reached by some choice of9.
Problem 2.2.3. Evaluate J; eVxdx. Show that Jo"" e-x•dx = r(i).
Problem 2.2.4. To get some familiarity with what kind ofmanipulations are legal, let us compute C ( R ), the circumference of a circle x2 + y2 = R2 • We will find that part that lies in the first quadrant and multiply by 4. Consider two points on the circle separated by (Ax, Ay). The arc length between them may be approximated by the pythagorean distance
The d 's are however not independent: If ( x, y) satisfies the equation for the circle, so must the displaced point. Show that this implies ~ = - i to leading order in the inflnitesimals. Consequently C(R) = f ds = 4f0R Jl + (i)2dx. Eliminate y in favor of x in the integrand and evaluate it by a trigonometric change of variables.
What would be the formula for the arc length of a general curve y -J(x) = 0?
Problem 2.2.5. Evaluate f0" 12 (~~~~:;).
Problem 2.2.6. Find the area enclosed between the curve y = x2 and the lines
Y = lxl
Problem 2.2.7. Evaluate J~ 2zdx.
Let us now consider the second trick of differentiating with respect to a parameter.
Consider
lo(a) = fooo e-az2dx (2.2.13)
This integral cannot be evaluated by any of the standard means. On the other hand if the integral in question had been
lt(a) = fooo e-a~~:2xdx (2.2.14) we could have changed to u = x2 and evaluated the integral.
Problem 2.2.8. Show that I 1 (a) = 2~.
Let us see how far we can get with Io(a). The notation itself tells us that the integral depends on just the parameter a. Its dependence on a can be found by scaling. In terms of u = .jO.x,
( ) 1 roo u2 1
Io a = Va Jo e- du = Vac (2.2.15)
where c is a constant independent of a. In the next chapter we will learn how to do this very important integral and find out that c = J/j-, so that
lo(a) = 2V ;;ã l{i (2.2.16)
Now it turns out that given Io(a), we c:an evaluate a whole family of related integrals by differentiating both sides of Eqn. (2.2.13) with respect to a. Doing
this once we find da
_.2_ E.
4a V-;;,
100 ae-a:z:2
= - - d x
0 aa
= 100 ( -x2)e-az2 dx so that 100{ 2) -az2d _ 1 ~
- X e X - - -.
o 4a a
This trick can be used to evaluate
ln(a) = 100 (x")e-a:z:2dx
(2.2.17) (2.2.18) (2.2.19)
(2.2.20) for n even. If n is odd, we can switch to x2 as the integration variable. The details are left to the following exercise.
Problem 2.2.9. Evaluate l3(a), l4(a).
This business of going inside an integral and differentiating with respect to a parameter like a seems to make some students a bit uneasy. Let us see why this is perfectly legal in a general case where
F(a) = J J(x,a)dx (2.2.21)
where it is assumed the limits are independent of a. Consider the following se- quence of operations
F(a +~a) j f(x, a+ ~a)dx (2.2.22)
F(a +~a)-F(a) = J f(x, a+ ~a)dx- J f(x, a)dx (2.2.23)
= jU(x,a +~a)-f(x,a))t;lx (2.2.24) Dividing both sides by ~a and taking the limit we get the desired result
dF(a) = J 8f(x,a) dx.
da aa (2.2.25)
Sometimes we have to be more devious. Recall from Exercise (2.2.2.) that (2.2.26)
By taking the a derivative we can show that
(2.2.27) Suppose we only knew that
(2.2.28) and wanted to evaluate the integral with {x2 + 1)2 in the denominator. We have no parameter to differentiate. Then we must introduce one! First we view the given integral as
roo dx
.fo x2 + a2
evaluated at a = 1. Next we can change to u = ax in Eqn. (2.2.28) and deduce Eqn. (2.2.26). Thereafter we proceed as before in taking parametric derivatives.
At the end we set a = 1. The notion of embedding a zero parameter problem into a parametrized family is very powerful and has many uses.
Problem 2.2.10. Consider
11 t -1
I -- - -
0 In t ã
Think of the tint- 1 as the a= 1 limit oft4 • Let I(a) be the corresponding integral. Take the a derivative of both sides (using t4 = e41"t) and evaluate dl fda by evaluating the corresponding integral by inspection. Given dl Ida obtain I by performing the indefinite integral of both sides with respect to a. Determine the constant of integration using your knowledge of 1(0). Show that the original integral equals In 2.
Problem 2.2.11. Given
1o 00 e-ax sin kxdx = -a 2- -+ k k 2 ,
evaluate J000 xe-ax sin kxdx and f0oc xe-ax cos kxdx.