# What integration technique to use?

1. Dec 24, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
What technique should I use to evaluate

$$\int_0^1 dx \frac{x^{m-1}}{\ln x}$$

where m is a positive integer

?

2. Relevant equations

3. The attempt at a solution

2. Dec 24, 2007

### rocomath

$$x^{m-1}=x^{m}\times x^{-1}$$

3. Dec 24, 2007

### Gib Z

Don't go around trying to find the anti-derivative of that in elementary functions, the indefinite integral evaluates to $$Ei (m \ln x) + C, \frac{d}{dx}C = 0$$ where Ei(x) is the exponential integral. This result follows because after the substitution u= ln x, the original integral becomes $$\int^0_{-\infty} \frac{ (e^u)^m}{u} du$$.

Best you can do is to get an infinite series solution.

Last edited: Dec 24, 2007
4. Dec 25, 2007

### Gib Z

Ahh I think I've found out how to solve it, because those bounds looked quite nice =] Try Differentiation under the integral.

5. Dec 25, 2007

### ehrenfest

Yes. That works. My original question was a mistake, however. I wanted to use parameter differentiation to find

$$H(m) = \int_0^1 dx \frac{x^{m}}{\ln x}$$

So, I differentiated erroneously and got

$$H(m) = \int_0^1 dx (m-1)\frac{x^{m-1}}{\ln x}$$

$$H(m) = \int_0^1 dx x^{m}$$

By, the way, $$H(m) = \ln{m+1}$$.

6. Dec 25, 2007

### Gib Z

Ah no it's not as easy as that...You have forgotten to deal with the Constant associated with the integrals..

7. Dec 25, 2007

### Gib Z

Well I might as well give another clue, try looking at H(-1) to evaluate the constant. If my mind was a bit clear yesterday, I could have told you that; looking at the natural logs series between 0 and 1, that integral is......