How Do I Solve This Integral with Given Restrictions?

[Gianfelici]

Hello! I'm having some troubles with that integral:

$\int_0^{k} \frac{x^{\alpha}}{1 + \beta x} dx$

I've tried to think a lot on this but I've no idea how to solve it, so I hope someone could help me. Thank you!

 

[Adithyan]

Assuming 'k' is a variable, you could write f(k)= $\int_0^{k} \frac{x^{\alpha}}{1 + \beta x} dx$. Try finding f'(k) and solve the problem.

 

[micromass]

Hello! I'm having some troubles with that integral:

$\int_0^{k} \frac{x^{\alpha}}{1 + \beta x} dx$

I've tried to think a lot on this but I've no idea how to solve it, so I hope someone could help me. Thank you!

You can solve this with the hypergeometric function:
http://www.wolframalpha.com/input/?i=int+x^a/(1++b*x)dx
http://en.wikipedia.org/wiki/Hypergeometric_function

I don't think you can solve it without using such a function.

 

[HallsofIvy]

Assuming 'k' is a variable, you could write f(k)= $\int_0^{k} \frac{x^{\alpha}}{1 + \beta x} dx$. Try finding f'(k) and solve the problem.

f'(k) is, of course, $\frac{k^{\alpha}}{1+ \beta k}$ but I don't see how that helps find f(k).

 

[D H]

f'(k) is, of course, $\frac{k^{\alpha}}{1+ \beta k}$ but I don't see how that helps find f(k).

How it helps is that one can express the integral as an infinite series. Obviously, f(0)=0. All one needs to form the infinite series is f'(k), f''(k), and so on.

 

[Adithyan]

How it helps is that one can express the integral as an infinite series. Obviously, f(0)=0. All one needs to form the infinite series is f'(k), f''(k), and so on.

Yeah and this method requires that we know the values of a and b.

 

[D H]

No, it doesn't.

 

[Saitama]

The integral can be easily evaluated if $k\rightarrow \infty$ and has a nice result.

 

[Redbelly98]

The integral can be easily evaluated if $k\rightarrow \infty$ and has a nice result.

Well, it's also easy to evaluate the integral if either α→0 or β→0. But the OP asked for a general solution rather than a solution for k, α, or β approaching some limit.

Besides, as k→∞, the integral is only defined when α is between -1 and 0, right? For positive (or zero) α, the integral diverges as k→∞. For α<-1, there is a problem integrating in the region near x=0.

 

[Saitama]

Besides, as k→∞, the integral is only defined when α is between -1 and 0, right? For positive (or zero) α, the integral diverges as k→∞. For α<-1, there is a problem integrating in the region near x=0.

Of course there are restrictions. Another restriction is that $\beta >0$. With these, the result of the definite integral is:
$$\frac{-1}{\beta^{\alpha+1}}\frac{\pi}{\sin(\pi \alpha)}$$