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For $a^2<b$, is there an identity of evaluating the following integral?

$$\int_0^\infty \frac{dx}{x^2+2ax+b}$$

What about:

$$\int_0^\infty \frac{dx}{(x^2+2ax+b)^2}$$

My attempt is using partial fractions and completing the square, but I still failed to obtain a nice result.

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  • $\begingroup$ Note that your first one is trivial and finite for $a=b=+1$, but quite different if $-a=b=1$. $\endgroup$ – almagest Jul 2 '16 at 14:24
  • $\begingroup$ Note: making change of variables $x-c$ we see that your first problem is the same as finding an indefinite integral of $1/(x^2+2ax+b)$. But note that doing the integral $\int_{-\infty}^\infty$ will give you a simpler answer. $\endgroup$ – GEdgar Jul 2 '16 at 19:58
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Let the first integral be $I$ and the second one be $J$, then by putting $x=y-a$ and $y=z\sqrt{b-a^2}$ we have \begin{align} I(a,b)&=\int_0^{\infty}\frac{dx}{(x+a)^2+b-a^2}\\[10pt] &= \int_a^{\infty}\frac{dy}{y^2+b-a^2}\\[10pt] &=\frac{1}{\sqrt{b-a^2}}\int_{\large\frac{a}{\sqrt{b-a^2}}}^{\infty}\frac{dz}{z^2+1}\\[10pt] &=\frac{1}{\sqrt{b-a^2}}\left(\frac{\pi}{2}-\arctan\left(\frac{a}{\sqrt{b-a^2}}\right)\!\right)\\[15pt] \end{align} and the 2nd integral is just a first derivative of $I$ with respect to $b$ times $-1$ \begin{equation} \\[15pt]J(a,b)=-\frac{\partial I}{\partial b}=\frac{\partial }{\partial b}\left[\frac{1}{\sqrt{b-a^2}}\left(\arctan\left(\frac{a}{\sqrt{b-a^2}}\right)-\frac{\pi}{2}\right)\right]\\[10pt] \end{equation} Can you take it from here?

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  • $\begingroup$ I think it is: $J(a,b)=-\frac{\partial I}{\partial b}$ $\endgroup$ – YY Lam Jul 2 '16 at 14:52
  • $\begingroup$ @YYLam Yes, you're correct. Edited. $\endgroup$ – Sophie Agnesi Jul 2 '16 at 15:00
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    $\begingroup$ Thank you very much. Now I obtain: $\frac{a}{2 b \left(a^2-b\right)}+\frac{\pi }{4 \left(b-a^2\right)^{3/2}}-\frac{\tan ^{-1}\left(\frac{a}{\sqrt{b-a^2}}\right)}{2 \left(b-a^2\right)^{3/2}}$ for $J(a,b)$ $\endgroup$ – YY Lam Jul 2 '16 at 15:05
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For the first one:

$$(x+a)^2=x^2+2ax+a^2$$

So

$$(x+a)^2+b-a^2=x^2+2ax+b$$

$$=(b-a^2)\left(\frac{(x+a)^2}{b-a^2}+1 \right)$$

$$=(b-a^2) \left( \left( \frac{x+a}{\sqrt{b-a^2}} \right)^2+1 \right)$$

For $b-a^2 \neq 0$. In which case, for evaluating the integral, we enforce the substitution $\tan {u}=\frac{x+a}{\sqrt{b-a^2}}$ and proceed from there. Or the substitution $t=\frac{x+a}{\sqrt{b-a^2}}$, $\sqrt{b-a^2} dt=dx$:

$$\int_{0}^{\infty} \frac{1}{(b-a^2) \left( \left( \frac{x+a}{\sqrt{b-a^2}} \right)^2+1 \right)} dx$$

$$=\frac{\sqrt{b-a^2}}{b-a^2} \int_{\frac{a}{\sqrt{b-a^2}}}^{\infty} \frac{1}{t^2+1} dt$$

$$=\frac{\sqrt{b-a^2}}{b-a^2}\left(\frac{\pi}{2}-\arctan (\frac{a}{\sqrt{b-a^2}}) \right)$$

$$=\frac{\text{arccot} (\frac{a}{\sqrt{b-a^2}})}{\sqrt{b-a^2}}$$

The second integral can be found by the method already mentioned, i.e. finding:

$$\frac{\partial}{\partial b} \left(-\frac{\text{arccot} (\frac{a}{\sqrt{b-a^2}})}{\sqrt{b-a^2}} \right)$$

In fact if we take:

$$\begin{equation*} I_n = \int^{\infty}_{0}\frac{1}{(x^2 + 2ax + b)^n} \mathrm{d}x\end{equation*}$$

Like zain did,

Then for $n \geq 2$:

$$I_n=\frac{\partial^{n-1}}{\partial b} \left(\frac{(-1)^{n-1}}{(n-1)!}\frac{\text{arccot} (\frac{a}{\sqrt{b-a^2}})}{\sqrt{b-a^2}} \right)$$

Because:

$$\frac{\partial^{n}}{\partial x} \frac{1}{x+c}=\frac{(-1)^nn!}{(x+c)^{n+1}}$$

Thus:

$$I_n(a,b)=\frac{(-1)^{n-1}}{(n-1)!}\frac{\partial^{n-1}}{\partial b_0} \left(\frac{\text{arccot} (\frac{a_0}{\sqrt{b_0-a_0^2}})}{\sqrt{b_0-a_0^2}} \right) \biggr \rvert_{(a,b)}$$

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