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Evaluation of $\displaystyle \int^{\infty}_{0}\frac{x}{(x+\alpha)(x^4+x^2+1)}dx~,~~\alpha>0$

Let \begin{align} I &=\int^{\infty}_{0}\frac{x}{(x+\alpha)(x^4+x^2+1)}dx \\ &=\int^{\infty}_{0}\frac{x}{(x+\alpha)(x^2-x+1)(x^2+x+1)}dx \end{align}

Although we can solve it as

$$\frac{x}{(x+a)(x^2-x+1)(x^2+x+1)}=\frac{A}{x+\alpha}+\frac{Bx+C}{x^2-x+1}+\frac{Dx+E}{x^2+x+1}$$

\begin{align} I &=\frac{1}{2}\int \bigg[\frac{(x^2+x+1)-(x^2-x+1)}{(x+\alpha)(x^2-x+1)(x^2+x+1)}\bigg]dx \\ &=\frac{1}{2}\int\frac{1}{(x+\alpha)(x^2-x+1)}dx-\frac{1}{2}\int\frac{1}{(x+\alpha)(x^2+x+1)}dx \end{align}

But it is very lengthy way

Can we solve it other then Partial Fraction Method?

Please have a look on this problem. Thanks.

Lee David Chung Lin
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    You can try solving it using the residue theorem, but partial fractions is quite easy and you are already on your way, so I would just use partial fractions unless you are familiar with residue theorem. – Abhishek Vangipuram Feb 10 '20 at 14:36

1 Answers1

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Continue with

$$\displaystyle I =\frac{1}{2}\int_0^\infty\frac{dx}{(x+\alpha)(x^2-x+1)}-\frac{1}{2}\int_0^\infty\frac{dx}{(x+\alpha)(x^2+x+1)}=\frac12I_1 - \frac12I_2 $$ where, $$I_1 =\int_0^\infty\frac{dx}{(x+\alpha)(x^2-x+1)}$$ $$=\frac1{a^2+a+1}\int_0^\infty\left(\frac{1}{x+\alpha}- -\frac12\frac{2x-1}{x^2-x+1}+\frac12\frac{1+2a}{x^2-x+1}\right)dx $$ $$=\frac1{a^2+a+1}\left( \ln\frac{x+\alpha}{\sqrt{x^2-x+1}} +\frac{1+2a}{\sqrt3}\tan^{-1} \frac{2x-1}{\sqrt3}\right)\bigg|_0^\infty $$

As a result, $$I_1=\frac{-\ln \alpha + \frac{2\pi}{3\sqrt3}(2a+1)}{a^2+a+1}$$

Similarly,

$$I_2 =\frac{-\ln a+ \frac{\pi}{3\sqrt3}(2a-1)}{a^2-a+1}$$

Thus,

$$I=\frac12I_1 - \frac12I_2 = \frac{a\ln a}{a^4+a^2+1} +\frac{\pi}{6\sqrt3}\frac{2a^3-3a^2+a+3}{a^4+a^2+1}$$

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