Chapter 15.1, Problem 53E

To determine

(a)

The mathematical proof of the statement stating that using L'Hopital Rule, $f(x)=\frac{e^{-x}-e^{-ax}}{x}$, though not defined at x = 0, can be made continuous by assigning the value $f(0)=a-1$

Answer to Problem 53E

Solution: The mathematical proof of the statement stating that using L'Hopital Rule, $f(x)=\frac{e^{-x}-e^{-ax}}{x}$, though not defined at x = 0, can be made continuous by assigning the value $f(0)=a-1$is derived.

Explanation of Solution

Explanation:

Given: $f(x)=\frac{e^{-x}-e^{-ax}}{x}$ for a>0

Calculation:

Step 1: The function $f(x)=\frac{e^{-x}-e^{-ax}}{x}$,$f(0)=a-1$ is continuous if ${\lim }_{x\to 0}f(x)=f(0)=a-1$.

Step 2: We verify this limit using L'Hopital Rule:

$\underset{x\to 0}{\lim }\frac{e^{-x}-e^{-ax}}{x}=\underset{x\to 0}{\lim }\frac{-e^{-x}+a{e}^{-ax}}{1}=-1+a=a-1$

Therefore, f is continuous.

Conclusion: The statement stating that using L'Hopital Rule, $f(x)=\frac{e^{-x}-e^{-ax}}{x}$, though not defined at x = 0, can be made continuous by assigning the value $f(0)=a-1$ is mathematically proved.

To determine

(b)

The mathematical proof of the statement stating that $|f(x)|\le e^{-x}+e^{-ax}$ for $x>1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem

Answer to Problem 53E

Solution: Both the mathematical proof of the statement stating that $|f(x)|\le e^{-x}+e^{-ax}$ for $x>1$ using triangle inequality and the mathematical proof of the statement stating that I(a) converges by applying the Comparison Theorem are derived.

Explanation of Solution

Given: $x>1$, $I(a)={\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{e^{-x}-e^{-ax}}{x}dx}$, $f(x)=\frac{e^{-x}-e^{-ax}}{x}$, a > 0

Calculation:

Step 1: We now show that the following integral converges:

$I(a)={\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{e^{-x}-e^{-ax}}{x}dx}$ (a>0)

Since, $e^{-x}-e^{-ax}<e^{-x}+e^{-ax}$ then $\frac{e^{-x}-e^{-ax}}{x}<\frac{e^{-x}+e^{-ax}}{x}$ for x > 0

If x > 1 we have,

$\frac{e^{-x}-e^{-ax}}{x}<\frac{e^{-x}+e^{-ax}}{x}<e^{-x}+e^{-ax}$

That is for x > 1

$f(x)<e^{-x}+e^{-ax}$

Step 2: Also, since $e^{-ax}-e^{-x}<e^{-ax}+e^{-x}$ we have for x > 1

$\frac{e^{-ax}-e^{-x}}{x}<\frac{e^{-ax}-e^{-x}}{x}<e^{-ax}+e^{-x}$

Thus, we get

$-f(x)<e^{-x}+e^{-ax}$

Step 3: Hence, from Step 1 and Step 2, we get

$0\le |f(x)|\le e^{-x}+e^{-ax}$

Step 4: We now show that the integral of the right hand side converges:

$\begin{array}{l}{\displaystyle \underset{0}{\overset{\infty }{\int }}(e^{-x}+e^{-ax})dx=\underset{R\to \infty }{\lim }}{\displaystyle \underset{0}{\overset{R}{\int }}(e^{-x}+e^{-ax})dx}=\underset{R\to \infty }{\lim }\left(-e^{-x}-{\frac{e^{-ax}}{a}|}_{x=0}^R\right)\\ =\underset{R\to \infty }{\lim }\left(e^{-R}-\frac{e^{-aR}}{a}+e^0+\frac{e^0}{a}\right)=\underset{R\to \infty }{\lim }\left(e^{-R}-\frac{e^{-aR}}{a}+1+\frac{1}{a}\right)=1+\frac{1}{a}\end{array}$

Since the integral converges, we conclude from Step 3 and the Comparison Test for Improper Integral that

${\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{(e^{-x}+e^{-ax})}{x}dx}$ also converges for a > 0.

Conclusion: Both the statement stating that $|f(x)|\le e^{-x}+e^{-ax}$ for $x>1$ using triangle inequality and the statement stating that I(a) converges by applying the Comparison Theorem are mathematically proved.

To determine

(c)

The mathematical proof of the equation $I(a)={\displaystyle \underset{0}{\overset{\infty }{\int }}{\displaystyle \underset{1}{\overset{a}{\int }}{e}^{-xy}dydx}}$

Answer to Problem 53E

Solution: The mathematical proof of the equation $I(a)={\displaystyle \underset{0}{\overset{\infty }{\int }}{\displaystyle \underset{1}{\overset{a}{\int }}{e}^{-xy}dydx}}$ is derived

Explanation of Solution

Given: $I(a)={\displaystyle \underset{0}{\overset{\infty }{\int }}{\displaystyle \underset{1}{\overset{a}{\int }}{e}^{-xy}dydx}}$, a>0

Calculation:

Step 1: We compute the inner integral with respect to y:

${\displaystyle \underset{1}{\overset{a}{\int }}{e}^{-xy}dy=-\frac{1}{x}}{e^{-xy}|}_{y=1}^a=-\frac{1}{x}(e^{-xa}-e^{-x.1})=\frac{e^{-x}-e^{-ax}}{x}$

Step 2: Hence,

$I(a)={\displaystyle \underset{0}{\overset{\infty }{\int }}{\displaystyle \underset{1}{\overset{a}{\int }}{e}^{-xy}dydx}}={\displaystyle \underset{0}{\overset{\infty }{\int }}\left({\displaystyle \underset{1}{\overset{a}{\int }}{e}^{-xy}dy}\right)}dx={\displaystyle \underset{0}{\overset{\infty }{\int }}\left(\frac{e^{-x}-e^{-ax}}{x}\right)}dx=I(a)$

Conclusion: The equation $I(a)={\displaystyle \underset{0}{\overset{\infty }{\int }}{\displaystyle \underset{1}{\overset{a}{\int }}{e}^{-xy}dydx}}$ is mathematically proved.

To determine

(d)

By interchanging the order of integration, the mathematical proof of the equation

$I(a)=\ln a-\underset{T\to \infty }{\lim }{\displaystyle \underset{1}{\overset{a}{\int }}\frac{e^{-Ty}}{y}dy}$

Answer to Problem 53E

Solution: By interchanging the order of integration, the mathematical proof of the equation

$I(a)=\ln a-\underset{T\to \infty }{\lim }{\displaystyle \underset{1}{\overset{a}{\int }}\frac{e^{-Ty}}{y}dy}$ is derived.

Explanation of Solution

Given: $I(a)={\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{1}{\overset{a}{\int }}{e}^{-xy}dydx}}$, a>0

Calculation:

Step 1: By the definition of the improper integral,

$I(a)=\underset{T\to \infty }{\lim }{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{1}{\overset{a}{\int }}{e}^{-xy}dy}}dx$

Step 2: We compute the double integral. Using Fubini's Theorem we may compute the iterated integral using the reversed order of integration. That is,

$\begin{array}{l}{\displaystyle \underset{0}{\overset{T}{\int }}{\displaystyle \underset{1}{\overset{a}{\int }}{e}^{-xy}dydx}}={\displaystyle \underset{1}{\overset{a}{\int }}{\displaystyle \underset{0}{\overset{T}{\int }}{e}^{-xy}dxdy={\displaystyle \underset{1}{\overset{a}{\int }}\left({\displaystyle \underset{0}{\overset{T}{\int }}{e}^{-xy}dx}\right)}}}dy={\displaystyle \underset{1}{\overset{a}{\int }}\left({-\frac{1}{y}{e}^{-xy}|}_{x=0}^T\right)}dy\\ ={\displaystyle \underset{1}{\overset{a}{\int }}\left(-\frac{1}{y}(e^{-Ty}-e^{-0.y})\right)dy={\displaystyle \underset{1}{\overset{a}{\int }}\frac{1-e^{-Ty}}{y}}dy={\displaystyle \underset{1}{\overset{a}{\int }}\frac{dy}{y}-{\displaystyle \underset{1}{\overset{a}{\int }}\frac{e^{-Ty}}{y}}dy}}\\ ={\ln y|}_1^a-{\displaystyle \underset{1}{\overset{a}{\int }}\frac{e^{-Ty}}{y}}dy=\ln a-\ln 1-{\displaystyle \underset{1}{\overset{a}{\int }}\frac{e^{-Ty}}{y}}dy=\ln a-{\displaystyle \underset{1}{\overset{a}{\int }}\frac{e^{-Ty}}{y}}dy\end{array}$

Combining with Step 1, we get,

$I(a)=\ln a-\underset{T\to \infty }{\lim }{\displaystyle \underset{1}{\overset{a}{\int }}\frac{e^{-Ty}}{y}dy}$

Conclusion: By interchanging the order of integration, the equation $I(a)=\ln a-\underset{T\to \infty }{\lim }{\displaystyle \underset{1}{\overset{a}{\int }}\frac{e^{-Ty}}{y}dy}$ is mathematically proved.

To determine

(e)

The mathematical proof of the statement stating the limit in $I(a)=\ln a-\underset{T\to \infty }{\lim }{\displaystyle \underset{1}{\overset{a}{\int }}\frac{e^{-Ty}}{y}dy}$ is zero by using the Comparison Theorem.

Answer to Problem 53E

Solution: The mathematical proof of the statement stating the limit in $I(a)=\ln a-\underset{T\to \infty }{\lim }{\displaystyle \underset{1}{\overset{a}{\int }}\frac{e^{-Ty}}{y}dy}$ is zero by using the Comparison Theorem is derived.

Explanation of Solution

Given: $I(a)=\ln a-\underset{T\to \infty }{\lim }{\displaystyle \underset{1}{\overset{a}{\int }}\frac{e^{-Ty}}{y}dy}$, a>0

Calculation:

Step 1: We consider the following possible cases:

Case 1: $a\ge 1$ then in the interval of integration $y\ge 1$. As $T\to \infty$, we may assume that T > 0

Thus,

$\frac{e^{-Ty}}{y}\le \frac{e^{-T.1}}{1}=e^{-T}$

Hence,

$0\le {\displaystyle \underset{1}{\overset{a}{\int }}\frac{e^{-Ty}}{y}dy\le }{\displaystyle \underset{1}{\overset{a}{\int }}{e}^{-T}dy}=e^{-T}(a-1)$

By the limit $\underset{T\to \infty }{\lim }{e}^{-T}(a-1)=0$ and the Squeeze Theorem, we conclude that,

$\underset{T\to \infty }{\lim }{\displaystyle \underset{1}{\overset{a}{\int }}\frac{e^{-Ty}}{y}dy}=0$

Case 2: $0<a<1.$

Then,

${\displaystyle \underset{1}{\overset{a}{\int }}\frac{e^{-Ty}}{y}dy=-}{\displaystyle \underset{a}{\overset{1}{\int }}\frac{e^{-Ty}}{y}dy}$

and in the interval of integration $a\le y\le 1$, therefore

$\frac{e^{-Ty}}{y}\le \frac{e^{-Ta}}{a}$(the function $\frac{e^{-Ty}}{y}$is decreasing).

Hence,

$0\le {\displaystyle \underset{a}{\overset{1}{\int }}\frac{e^{-Ty}}{y}dy\le }{\displaystyle \underset{a}{\overset{1}{\int }}\frac{e^{-Ta}}{a}dy}=\frac{1-a}{a}{e}^{-Ta}$

By the limit $\underset{T\to \infty }{\lim }\frac{1-a}{a}{e}^{-Ta}=0$ and the Squeeze Theorem, we conclude that

$\underset{T\to \infty }{\lim }{\displaystyle \underset{1}{\overset{a}{\int }}\frac{e^{-Ty}}{y}=-}\underset{T\to \infty }{\lim }{\displaystyle \underset{a}{\overset{1}{\int }}\frac{e^{-Ty}}{y}=0}$

We thus showed that for all a > 0, $\underset{T\to \infty }{\lim }{\displaystyle \underset{1}{\overset{a}{\int }}\frac{e^{-Ty}}{y}=0}$

Conclusion: The statement stating the limit in $I(a)=\ln a-\underset{T\to \infty }{\lim }{\displaystyle \underset{1}{\overset{a}{\int }}\frac{e^{-Ty}}{y}dy}$ is zero by using the Comparison Theorem is mathematically proved.