MHB Integral of Exponential Fractions with Positive Parameters

[polygamma]

Show that for positive parameters $a$, $b$, and $c$,

$$ \int_{0}^{\infty} \left( \frac{e^{-ax}-e^{-bx}}{x^{2}} + (a-b) \frac{e^{-cx}}{x} \ \right) \ dx = b-a + a \ln \left(\frac{a}{c} \right) - b \ln \left(\frac{b}{c} \right)$$

 

[alyafey22]

[sp]

$$ F = \int_{0}^{\infty} \left( \frac{e^{-ax}-e^{-bx}}{x^{2}} + (a-b) \frac{e^{-cx}}{x} \ \right) \ dx$$

$$F_a =\int_{0}^{\infty} \frac{e^{-cx}-e^{-ax}}{x}\ dx = \ln \left( \frac{a}{c}\right) $$

$$F= a \ln \left( \frac{a}{c}\right) - a + C $$

$$F_b = C_b$$

$$F_b = \int_{0}^{\infty} \frac{e^{-bx}-e^{-cx}}{x} dx = -\ln \left( \frac{b}{c}\right)$$

We have $$C_b = -\ln \left( \frac{b}{c}\right)$$ so

$$C = -b\ln \left( \frac{b}{c}\right)+b$$

$$F = b-a +a \ln \left( \frac{a}{c}\right)-b\ln \left( \frac{b}{c}\right)$$

[/sp]

 

[polygamma]

@ Zaid

I actually didn't even consider differentiating inside of the integral.

I took the boring approach of finding an antiderivative in terms of the exponential integral and then using the expansion of the exponential integral at $x=0$ to evaluate the antiderivative at the lower limit.

There is one thing about your evaluation that I'm not sure about, though.

When you integrated to find the constant $C$, how did you know that he constant of integration was zero?

Alternatively what you could do to find $C$ is to let $a=b$ in the original integral.

 

[alyafey22]

There is one thing about your evaluation that I'm not sure about, though.

When you integrated to find the constant $C$, how did you know that he constant of integration was zero?

Alternatively what you could do to find $C$ is to let $a=b$ in the original integral.

Yeah, I made the evaluations not clear but I actually considered that $C$ as a function of $b$ so I differentiated with respect to $b$. Again putting $a=b$ is by far much clearer.

Can you post your approach ?

 

[polygamma]

It's easier to work the upper incomplete gamma function.

Spoiler

\Gamma(0,a x) = -\text{Ei}(- ax) = \int_{ax}^{\infty} \frac{e^{-t}}{t} \ dt = \int_{x}^{\infty} \frac{e^{-au}}{u} \ duLet's first derive the expansion at $x=0$.

$$ \Gamma(0,x) = \int_{x}^{\infty} \frac{e^{-t}}{t} \ dt = \int_{1}^{\infty} \frac{e^{-t}}{t} \ dt - \int_{1}^{x} \frac{e^{-t}}{t} \ dt $$

$$ = \int_{1}^{x} \frac{1-e^{-t}}{t} \ dt - \ln x + \int_{1}^{\infty} \frac{e^{-t}}{t} \ dt$$

$$= -\ln x + \Big( -\int_{0}^{1} \frac{1-e^{-t}}{t} \ dt + \int_{1}^{\infty} \frac{e^{-t}}{t} \ dt \Big) + \int_{0}^{x} \frac{1-e^{-t}}{t} \ dt$$

$$ = - \ln x - \gamma - \int_{0}^{x} \frac{e^{-t}-1}{t} \ dt$$

$$ = - \ln x - \gamma \int_{0}^{x} \Big(-1 + \frac{t}{2} - \frac{t^{2}}{6} + \ldots \Big) \ dt$$

$$ = - \ln x - \gamma + x + \mathcal{O}(x^{2}) $$And it's not really needed here, but one can find an asymptotic expansion at $x= \infty$ by integrating by parts.

$$ \Gamma(0,x) \sim \frac{e^{-x}}{x} + \mathcal{O}\left( \frac{e^{-x}}{x^{2}} \right) \ \ x \to \infty$$Then

$$ \int \left( \frac{e^{-ax}-e^{-bx}}{x^{2}} + (a-b) \frac{e^{-cx}}{x} \ \right) \ dx = \int \frac{e^{-ax}-e^{-bx}}{x^{2}} \ dx + (a-b) \int \frac{e^{-cx}}{x} \ dx$$

$$ = -\frac{e^{-ax} -e^{-bx}}{x} + \int \frac{-ae^{-ax}+be^{-bx}}{x} \ dx -(a-b) \Gamma(0,cx)$$

$$ = \frac{e^{-bx} -e^{-ax}}{x} + a \Gamma(0,ax) - b \Gamma(0,bx) - (a-b) \Gamma(0,cx)+C$$$$ \int^{\infty}_{0} \left( \frac{e^{-ax}-e^{-bx}}{x^{2}} + (a-b) \frac{e^{-cx}}{x} \ \right) \ dx$$

$$ = 0 - \lim_{x \to 0} \left[ \frac{e^{-bx} -e^{-ax}}{x} + a \left(-\ln ax - \gamma \right) - b \left(-\ln bx - \gamma \right) -(a-b) \left(-\ln cx - \gamma \right) \right]$$

$$= - \lim_{x \to 0} \left[ \frac{e^{-bx} -e^{-ax}}{x} - a \ln \left(\frac{a}{c} \right) + b \left(\frac{b}{c} \right) \right]$$

$$ = b-a + a \ln \left(\frac{a}{c} \right) - b \ln \left(\frac{b}{c} \right) $$