# Calculate the value for the following improper integral

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 thats 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 thats I calculate limit and got $$\ln3-\ln2$$.

I am not sure if thats 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: $$\int_0^{\infty}=\int_0^{1}+\int_{1}^{3/2}+\int_{3/2}^{\infty} ?$$
• You have to do it by 2) as you said, because the asymptotic – Ahlfkushevich Mar 1 at 11:22
• @JoseSquare then, just to make it sure. when I have to solve improper integral, any integral. the first thing thats should be done is to find singularity points which we the function is not define. and then separate the integral, right? – John D Mar 1 at 11:26
• Yes your are right, whenever the integrand becomes infinity in some point you have to study the convergence of the integral in that point – Ahlfkushevich Mar 1 at 11:30
• You can use Fundamental Theorem of integral calculus only when your function is continuous! Your function has two singularities, thus you have to worry about them before considering the limit as $x \to \infty$. – Crostul Mar 1 at 11:31
• For future reference, Wolfram|Alpha is a great place to check your answers if you aren't sure you are right – lioness99a Mar 1 at 12:20

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.

• According to answer 2, in this case I should calculate the value and not just say if the integral convergent. Why is it enough to show only the integral from 0 to 1 and that`s all? – John D Mar 1 at 11:30
• Once again: $\int_0^{\infty}$ is convergent $\iff$ the integrals $\int_0^{1},\int_{1}^{3/2}$ and $\int_{3/2}^{\infty}$ are convergent. Conclusion: if one of the three integrals is divergent, then $\int_0^{\infty}$ isdivergent. – Fred Mar 1 at 11:39

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):

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

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

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

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):

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