# Integration by Differentiating with Respect to a Parameter

1. Sep 15, 2009

### Piamedes

1. The problem statement, all variables and given/known data

Consider

$$I = \int^{1}_{0} \frac{t-1}{ln t}$$

(No, the problem does not contain a dt at the end of the integral, but I'll assume its a typo)

Think of the t in t-1 as the a=1 limit of t^a. Let I(a) be the corresponding integral. Take the a derivative of both sides and evaluate dI/da by evaluating the corresponding integral by inspection. Given dI/da obtain I by performing the indefinite integral of both sides with respect to a. Determine the constant of integration using your knowledge of I(0). Show htat the original integral equals ln2.

My attempt at a solution

Assuming a=1

$$I = \int^{1}_{0} \frac{t^{a}-1}{ln t} dt$$

$$\frac{dI}{da} = \int^{1}_{0} \frac{t^{a}lnt}{ln t}dt$$

$$\frac{dI}{da} = \int^{1}_{0} t^{a}dt$$

And here is where I start to have issues. If I follow the book's instructions I fail to achieve anything:

$$\int dI =\int \int^{1}_{0} t^{a}dt da$$

$$I = C + \int^{1}_{0} \frac{t^{a}}{ln t} dt$$

If assume that for I(0) it means to substitute a=0 for a=1, then I(0)=0, and then I can't get any farther than that.

However by not following the book's instructions I think I get an actual answer. The process is the same until:

$$\frac{dI}{da} = \int^{1}_{0} t^{a}dt$$

Then I continue:

$$\frac{dI}{da} = \frac{t^{a+1}}{a+1} ]^{1}_{0}$$

$$\frac{dI}{da} = \frac{1^{a+1}}{a+1} - \frac{0^{a+1}}{a+1}$$

$$\frac{dI}{da} = \frac{1}{a+1}$$

$$dI = \frac{1}{a+1}da$$

$$\int dI = \int \frac{1}{a+1}da$$

$$I = C + ln(a+1)$$

But a=1, so:

$$I = C + ln(2)$$

If I follow the same logic with regard to I(0) as before, then C=0 and:

$$I = ln(2)$$

Which is what the question asked for. However I'm unsure if that is the proper way to solve for I(0), or even if the ways I integrated are proper?

So basically my question is does this look mathematically correct, both method and answer? If not, then in what other way should I go about solving the equation?

2. Sep 15, 2009

### Dick

Your second method is perfectly correct. And I think that's what the book's instructions are intended to convey.

3. Sep 15, 2009

Thanks