Calculate the value for the following improper integral

Question

I need to calculate the value of the following integral:

$$\int_0^{\infty}\frac{dx}{2x^2-5x+3}$$

Below is how I tried to solve it and I am not sure that`s the correct value.

$ \int_0^{\infty}\frac{dx}{2x^2-5x+3} = \int_0^{\infty}\frac{dx}{(2x-3)(x-1)} = \int_0^{\infty}\frac{2dx}{(2x-3)} - \int_0^{\infty}\frac{dx}{(x-1)} = \ln|2x-3|-\ln|x-1| $

After that`s I calculate limit and got $\ln3-\ln2$.

I am not sure if that`s the correct way to solve so my questions to the current integral as following:

1. If my way is not correct - please advice why and what is the correct method to solve it.
2. I see that there is asymptotic at $x=1$ and $x={3/2}$ so should we need to calculate integral as following: $ \int_0^{\infty}=\int_0^{1}+\int_{1}^{3/2}+\int_{3/2}^{\infty} ? $

Answer 1

Question 1: your way is not correct (reason: see my answer to question 2).

Question 2: $\int_0^{\infty}$ is convergent $ \iff$ the integrals $\int_0^{1},\int_{1}^{3/2}$ and $\int_{3/2}^{\infty} $ are all convergent.

Now show that $\int_0^{1}\frac{dx}{2x^2-5x+3}$ is divergent !

Conclusion: $\int_0^{\infty}\frac{dx}{2x^2-5x+3}$ is divergent.

Answer 2

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):

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

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

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

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So, in your case we have $\text{n}=0$, $\text{a}=2$, $\text{b}=-5$ and $\text{c}=3$ (but as said by other users your integral does not converge, but when it does it has the value):

$$\mathscr{I}_0\left(2,-5,3\right)=\mathcal{PV}\int_0^\infty\frac{1}{2x^2-5x+3}\space\text{d}x=$$ $$\lim_{\text{n}\to0}\left(-1\right)^{1+\text{n}}\cdot\left(\left(\frac{3}{2}\right)^\text{n}-1\right)\cdot\pi\cdot\csc\left(\text{n}\cdot\pi\right)=-\ln\left(\frac{3}{2}\right)\tag7$$