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The integral

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

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

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

or, differentiating

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

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

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

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

given that the two logarithms are calculable.

This would, for the initial example, give us

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