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Math V21B Section 7.8

Introduction

Please try this problem on your own before looking at my solution. There may be other ways to solve this; I make no claim that this method is the quickest, prettiest, or most efficient.

Problem 37

The original problem is $$I = \int_{ - 1}^0 {\frac{{{e^{1/x}}}} {{{x^3}}}\:dx} $$

The integrand is undefined at $x = 0$, so we conclude that this is likely to be a vertical asymptote, and that we have a Type II improper integral to solve. Because the region (interval) of integration lies immediately to the left of this asymptote, we need to use only one-sided (in this case, left-sided) limits to evaluate the behavior of the antiderivative near $x = 0$. Let us re-write the integral in the form $$I = \int_{ - 1}^0 {\frac{{{e^{1/x}}}} {1}\frac{1} {x}\frac{1} {{{x^2}}}\:dx} $$ which immediately suggests the $w$-substitution $$w = \frac{1} {x} = {x^{ - 1}};\;\;\;\;\;dw = - {x^{ - 2}}dx = - \frac{1} {{{x^2}}}dx.$$ Re-writing $I$ with this substitution gives $$I = \int_?^? {{e^w} \cdot w \cdot \left( { - dw} \right)} = - \int_?^? {w{e^w}\:dw} $$ where the question marks represent the new limits of integration in terms of $w$ rather than $x$, and the negative sign arises from taking the differential of $x^{ - 1}$. Normally, we would explicitly find numeric values for these limits, but since $w = \frac{1} {x}$ is undefined at $x = 0$, replacing the upper limit would be troublesome. So, as an alternative, we will temporarily employ the question marks as surrogate limits until we find a way to work around this issue.

The latter integral is a classic integration-by-parts problem (or we could "cheat" and look up the result in a table). Using the "LIPTE" mnemonic, we set $u$ equal to the algebraic function $w$, from which the rest of the substitutions immediately follow: $$u = w;\;\;\;\;\;du = dw;\;\;\;\;\;dv = {e^w}\:dw;\;\;\;\;\;v = {e^w}$$ Employing the $uv - \int{v\:du}$ paradigm, and not neglecting to specify limits in all required locations, we obtain $$I = - \left[ {\left. {w{e^w}} \right|_?^? - \int_?^? {{e^w}\:dw} } \right] = - \left[ {w{e^w} - {e^w}} \right]_?^? = - \left[ {\left({e^w}\right)\cdot\left( {w - 1} \right)} \right]_?^? = \left[ {\left({e^w}\right)\cdot\left( {1 - w} \right)} \right]_?^?$$ where the leading negative sign disappears in the last step as a result of exchanging the operands in the parenthesized subtraction.

We can now restore the original variable and the original limits of integration by replacing $w$ with $\frac{1}{x}$: $$I = \left[ {\left({e^{1/x}}\right)\cdot\left( {1 - \frac{1} {x}} \right)} \right]_{ - 1}^0 = \mathop {\lim }\limits_{t \to {0^ - }} \left[ {\left({e^{1/x}}\right)\cdot\left( {1 - \frac{1} {x}} \right)} \right]_{ - 1}^t $$ with the last step coming from the technical definition of the Type II improper integral. Again, note that the "lim" limit is left-sided; by contrast, if the limits of integration had been, say, 0 to 1, we would have used a right-sided limit.

To begin the process of evaluating the "lim" limit, let's plug in the $t$ and the $-1$ separately, then subtract. $$I = \mathop {\lim }\limits_{t \to {0^ - }} \left[ {\left( {{e^{1/t}}} \right) \cdot \left( {1 - \frac{1} {t}} \right) - \left( {{e^{1/( - 1)}}} \right) \cdot \left( {1 - \frac{1} { - 1}} \right)} \right] = \mathop {\lim }\limits_{t \to {0^ - }} \left[ {\left( {{e^{1/t}}} \right) \cdot \left( {1 - \frac{1} {t}} \right)} \right] - \left( {{e^{ - 1}}} \right) \cdot \left( {2} \right)$$ $$I = \mathop {\lim }\limits_{t \to {0^ - }} \left[ {\left( {{e^{1/t}}} \right) \cdot \left( {1 - \frac{1} {t}} \right)} \right] - \left( {\frac{2} {e}} \right)$$

To help us understand the behavior of the "lim" limit, let us select a number near (but to the left side of) $t = 0$. (Because of the restriction to $t$ values less than zero, we must make selections such as $t = -0.001$ or $t = -0.0001$ rather than $t = 0.001$ or $t = 0.0001$.)

At $t = -0.001$, we find that $${e^{1/t}} = {e^{1/ - 0.001}} = {e^{ - 1000}} = \frac{1} {{{e^{1000}}}} = \frac{1} {{{\text{huge}}}} \approx 0$$ whereas $$1 - \frac{1} {t} = 1 - \frac{1}{ - 0.001} = 1 - ( - 1000) = 1001 \approx {\text{huge}}$$ which means that the product inside the rectangular brackets is of the form $\left( {0 \cdot {\text{huge}}} \right)$ or $\left( {0 \cdot \infty } \right)$. This is an indeterminate form!

Indeterminate forms are often susceptible to solution using L'Hôpital's rule. However, this rule requires the limiting function to be in the form of a quotient, not a product (which is what we have). With a few steps of manipulation, we can convert this to a quotient of the form $\frac{\infty } {\infty }$.

First, substitute a variable $z = \frac{1} {t}$ into the limit; as $t$ approaches 0 from the left, $z$ will approach negative infinity. (The reason for the new variable is to make the L'Hôpital's-rule derivatives much easier to compute.) With this new variable, we obtain $$I = \mathop {\lim }\limits_{z \to {- \infty} } \left[ {\left( {e^z} \right) \cdot \left( {1 - z} \right)} \right] - \left( {\frac{2} {e}} \right)$$ Second, re-write $e^z$ as $\frac {1}{e^{-z}}$, giving $$I = \mathop {\lim }\limits_{z \to - \infty } \left[ {\left( {\frac{1}{{{e^{ - z}}}}} \right) \cdot \left( {1 - z} \right)} \right] - \left( {\frac{2}{e}} \right) = \mathop {\lim }\limits_{z \to - \infty } \left( {\frac{{1 - z}}{{{e^{ - z}}}}} \right) - \left( {\frac{2}{e}} \right)$$

This quotient is in the desired L'Hôpital form, so apply the rule by differentiating numerator and denominator separately: $$I = \mathop {\lim }\limits_{z \to - \infty } \left( {\frac{{ - 1}} {{ - {e^{ - z}}}}} \right) - \left( {\frac{2} {e}} \right) = \mathop {\lim }\limits_{z \to - \infty } \left( {\frac{1} {{{e^{ - z}}}}} \right) - \left( {\frac{2} {e}} \right) = \mathop {\lim }\limits_{z \to - \infty } \left( {{e^z}} \right) - \left( {\frac{2} {e}} \right) = 0 - \left( {\frac{2}{e}} \right) = \boxed{\displaystyle{ - \frac{2}{e}}}$$

We conclude that the original integral converges to $- \frac{2}{e}$. The result is negative because the graph of the integrand function lies below the $x$-axis on the interval $[-1,0)$.