The integral of (e^(-px)-e^(-qx))/x dx from 0 to infinity

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The integral

$\int\limits_0^\infty {\frac{{{e^{ - x}}}}{x}dx}$

will diverge. It will behave as 1/x as x approaches 0, and the integral of that function does indeed diverge. But, as we will see an integral like

$\int\limits_0^\infty {\frac{{{e^{ - x}} - {e^{ - 2x}}}}{x}dx}$

will converge. The problem is that the primitive function seems to be very hard to find, so a direct approach seems to be difficult. One way to solve this is to introduce an extra variable – or why not two – to instead try to solve the integral

$I = \int\limits_0^\infty {\frac{{{e^{ - px}} - {e^{ - qx}}}}{x}dx}$

To start with it seems like we have not simplified the problem, until we realize that, if we take the derivative with respect to, say, p, then a factor of x would pop out of the exponential function, and that would cancel against the x in the denominator. So without further ado, let’s do it. We have

$I = I(p) = \int\limits_0^\infty {\frac{{{e^{ - px}} - {e^{ - qx}}}}{x}dx}$

or, differentiating

$I'(p) = \int\limits_0^\infty {\frac{{ - x{e^{ - px}}}}{x}dx} = \int\limits_0^\infty { - {e^{ - px}}dx}$

When taking the derivative we have to remember that p is our variable, and x is seen as a constant. Next, we integrate. This gives us

$I'(p) = \left[ {\frac{{{e^{ - px}}}}{p}} \right]_0^\infty = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{e^{ - px}}}}{p}} \right) - \frac{{{e^{ - p \cdot 0}}}}{p} = - \frac{1}{p}$

Ok, so we have found I´(p) in a rather simple form. To get back to I(p) we need to integrate with respect to p again. This gives us

$I(p) = \int { - \frac{{dp}}{p} = - \ln (p) + C}$

Next we need to find the value of the constant. To do this we choose a value of p that is possible to find the integral of. If we choose p=q then the two exponential functions in the original integral would cancel to 0, and the integral would thus evaluate to 0. We thus have

$I(q) = - \ln (q) + C = 0$

And C=ln(q). This would finally give us

$I = \int\limits_0^\infty {\frac{{{e^{ - px}} - {e^{ - qx}}}}{x}dx} = - \ln (p) + ln(q) = ln\frac{q}{p}$

given that the two logarithms are calculable.

This would, for the initial example, give us

$\int\limits_0^\infty {\frac{{{e^{ - x}} - {e^{ - 2x}}}}{x}dx = \ln (2/1) = \ln(2)}$

Up a level : Integrals
Previous page : Differentiation under the integral sign - Leibniz integral rule
Next page : The integral of sin(x)/x dx from -infinity to infinity

Last modified: Dec 28, 2023 @ 14:47