Finding a difficult integral [closed]

Good afternoon, I need to find:

$$\int_0^\infty\frac{x^n}{ax^2+bx+c}\space dx$$

I've no idea about any clue what so ever.

Solution 1:

Well, we have:

$$\mathscr{I}_{\space\text{n}}\left(\text{a},\text{b},\text{c}\right):=\int_0^\infty\frac{x^\text{n}}{\text{a}\cdot x^2+\text{b}\cdot x+\text{c}}\space\text{d}x\tag1$$

Using the Laplace transform we can write:

• When $\Re\left(\text{s}\right)>0\space\wedge\space\Re\left(\text{n}\right)>-1$:

$$\mathscr{L}_x\left[x^\text{n}\right]_{\left(\text{s}\right)}=\frac{\Gamma\left(1+\text{n}\right)}{\text{s}^{1+\text{n}}}\tag2$$

• When $\Re\left(\text{s}\right)>\Re\left(\text{z}_-\right)\space\wedge\space\Re\left(\text{z}_+\right)<\Re\left(\text{s}\right)$:

$$\mathscr{L}_x^{-1}\left[\frac{1}{\text{a}\cdot x^2+\text{b}\cdot x+\text{c}}\right]_{\left(\text{s}\right)}=\frac{\exp\left(\text{s}\cdot\text{z}_+\right)-\exp\left(\text{s}\cdot\text{z}_-\right)}{\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}\tag3$$

Where $\text{z}_{\pm}$ are the roots of $\text{a}\cdot x^2+\text{b}\cdot x+\text{c}$

Using the 'evaluating integrals over the positive real axis':

$$\mathscr{I}_{\space\text{n}}\left(\text{a},\text{b},\text{c}\right)=\int_0^\infty\frac{\Gamma\left(1+\text{n}\right)}{\text{s}^{1+\text{n}}}\cdot\frac{\exp\left(\text{s}\cdot\text{z}_+\right)-\exp\left(\text{s}\cdot\text{z}_-\right)}{\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}\space\text{d}\text{s}=$$

$$\frac{\Gamma\left(1+\text{n}\right)}{\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}\cdot\int_0^\infty\frac{\exp\left(\text{s}\cdot\text{z}_+\right)-\exp\left(\text{s}\cdot\text{z}_-\right)}{\text{s}^{1+\text{n}}}\space\text{d}\text{s}=$$

$$\frac{\Gamma\left(1+\text{n}\right)}{\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}\cdot\left\{\int_0^\infty\frac{\exp\left(\text{s}\cdot\text{z}_+\right)}{\text{s}^{1+\text{n}}}\space\text{d}\text{s}-\int_0^\infty\frac{\exp\left(\text{s}\cdot\text{z}_-\right)}{\text{s}^{1+\text{n}}}\space\text{d}\text{s}\right\}\tag4$$

Now, we need to look at:

• When $\Re\left(\text{z}_+\right)<0\space\wedge\space\Re\left(\text{n}\right)<0$

$$\int_0^\infty\frac{\exp\left(\text{s}\cdot\text{z}_+\right)}{\text{s}^{1+\text{n}}}\space\text{d}\text{s}=\left(-\text{z}_+\right)^\text{n}\cdot\Gamma\left(-\text{n}\right)\tag5$$

• When $\Re\left(\text{z}_-\right)<0\space\wedge\space\Re\left(\text{n}\right)<0$

$$\int_0^\infty\frac{\exp\left(\text{s}\cdot\text{z}_-\right)}{\text{s}^{1+\text{n}}}\space\text{d}\text{s}=\left(-\text{z}_-\right)^\text{n}\cdot\Gamma\left(-\text{n}\right)\tag6$$

So, we end up with (using the reflection formula):

$$\mathscr{I}_{\space\text{n}}\left(\text{a},\text{b},\text{c}\right)=\frac{\Gamma\left(1+\text{n}\right)}{\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}\cdot\left(\left(-\text{z}_+\right)^\text{n}\cdot\Gamma\left(-\text{n}\right)-\left(-\text{z}_-\right)^\text{n}\cdot\Gamma\left(-\text{n}\right)\right)=$$

$$\frac{\Gamma\left(-\text{n}\right)\cdot\Gamma\left(1+\text{n}\right)}{\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}\cdot\left(\left(-\text{z}_+\right)^\text{n}-\left(-\text{z}_-\right)^\text{n}\right)=$$

$$\frac{\pi\cdot\csc\left(\text{n}\cdot\pi\right)}{\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}\cdot\left(\left(-\text{z}_-\right)^\text{n}-\left(-\text{z}_+\right)^\text{n}\right)=$$ $$\left(-1\right)^\text{n}\cdot\frac{\pi\cdot\csc\left(\text{n}\cdot\pi\right)}{\sqrt{\text{b}^2-4\cdot\text{a}\cdot\text{c}}}\cdot\left(\text{z}_-^\text{n}-\text{z}_+^\text{n}\right)\tag6$$

As an example $\text{a}=\text{b}=1,\text{c}=0$:

$$\mathscr{I}_{\space\text{n}}\left(1,1,0\right)=\int_0^\infty\frac{x^\text{n}}{1\cdot x^2+1\cdot x+0}\space\text{d}x=\pi\cdot\csc\left(\text{n}\cdot\pi\right)\tag7$$

And when $\text{a}=\frac{\text{b}^2}{4\cdot\text{c}}$:

$$\mathscr{I}_{\space\text{n}}\left(\frac{\text{b}^2}{4\cdot\text{c}},\text{b},\text{c}\right)=\int_0^\infty\frac{x^\text{n}}{\frac{\text{b}^2}{4\cdot\text{c}}\cdot x^2+\text{b}\cdot x+\text{c}}\space\text{d}x=$$

$$\frac{2^{1+\text{n}}\cdot\left(\frac{\text{c}}{\text{d}}\right)^\text{n}\cdot\text{n}\cdot\pi\cdot\csc\left(\text{n}\cdot\pi\right)}{\text{b}}\tag8$$