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

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

will diverge. It will basically 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 an direct aprach 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 the 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: Jun 17, 2018 @ 19:20