# Find $\lim\limits_{t \to \infty} \int_{0}^{t} \frac{\mathrm dx}{(x^2+a^2)(x^2+b^2)(x^2+c^2)}$

Find $$\lim_{t \to \infty} \int_{0}^{t} \frac{1}{(x^2+a^2)(x^2+b^2)(x^2+c^2)}\mathrm dx$$ where $$a,b,c$$ are strict positive and dinstinct real numbers.

I know it should be something with arctangent but I don't know how to get there. Can somebody give me some tips, please?

• Do you mean limit as $\color{red}{t}\to \infty$, and integrating with respect to $x$? If so, one possibility could be to consider the partial fraction decomposition of $\frac{1}{\left(X+ a^2\right)\left(X+ b^2\right)\left(X+ c^2\right)}$ and replace $X$ with $x^2$ afterwards. – Minus One-Twelfth Mar 1 at 8:42
• Consider partial fractions – NoChance Mar 1 at 9:07

Using partial fraction decomposition, we have that \begin{align*} \frac1{(x^2+a^2)(x^2+b^2)(x^2+c^2)}&=\frac1{c^2-a^2}\left(\frac1{(x^2+a^2)(x^2+b^2)}-\frac1{(x^2+b^2)(x^2+c^2)}\right)\\&=\frac1{c^2-a^2}\left(\frac{1}{b^2-a^2}\left(\frac1{x^2+a^2}-\frac1{x^2+b^2}\right)\right)\\&-\frac1{c^2-a^2}\left(\frac{1}{c^2-b^2}\left(\frac1{x^2+b^2}-\frac1{x^2+c^2}\right)\right). \end{align*} Using $$\lim_{t\to\infty}\int_0^t \frac{1}{x^2+\alpha^2}\mathrm dx=\frac{\pi}{2\alpha}$$, $$\ \alpha>0$$, we can obtain the integral:

\begin{align*} I&=\frac1{c^2-a^2}\frac1{b^2-a^2}\left(\frac1{a}-\frac1b\right)\frac{\pi}2-\frac1{c^2-a^2}\frac1{c^2-b^2}\left(\frac1{b}-\frac1c\right)\frac{\pi}2 \\&=\frac1{c^2-a^2}\frac{\pi}{2ab(a+b)}-\frac1{c^2-a^2}\frac{\pi}{2bc(b+c)}\\ &=\frac{\pi(a+b+c)}{2abc(a+b)(b+c)(c+a)}. \end{align*}

Hints

Method 1: Partial fractions Since $$a, b, c$$ are positive and distinct, a straightforward application of the method of partial fractions gives that the integrand is $$\sum \frac{1}{(a^2 - b^2)(a^2 - c^2)} \frac{1}{x^2 + a^2},$$ where the sum is over the cyclic permutations of the parameters $$(a, b, c)$$. By linearity we can write the integral as a linear combination of integrals of the form $$\int_0^{\infty} \frac{dx}{x^2 + k^2} ,$$ and like you say, these can be evaluated with the usual antiderivative formula $$\int \frac{dx}{x^2 + k^2} = \frac{1}{k} \arctan \frac{x}{k} + C .$$

Method 2: Contour integration Since the integrand in even, it coincides with $$\frac{1}{2} \int_{-\infty}^\infty \frac{dx}{(x^2 + a^2) (x^2 + b^2) (x^2 + c^2)}.$$ This suggests integrating over a semicircle $$\Gamma_R$$ of radius $$R$$ with diameter on the real axis and centered at zero, say, in the upper half-plane. For large $$R$$, the value of the integral over the circular arc is $$O(R^{-5})$$, so in the limit its contribution to the integral is zero and hence $$\frac{1}{2} \int_{-\infty}^\infty \frac{dx}{(x^2 + a^2) (x^2 + b^2) (x^2 + c^2)} = \lim_{R \to \infty} \frac{1}{2} \oint_{\Gamma_R} \frac{dz}{(z^2 + a^2) (z^2 + b^2) (z^2 + c^2)} ,$$ but (for large $$R$$, and since $$a, b, c$$ are positive and distinct) the poles inside $$\Gamma_R$$ are simple poles at $$ia, ib, ic$$, so we can apply the residue formula, which gives that the integral is $$\frac{1}{2} \cdot 2 \pi i (\operatorname{Res}(f, ia) + \operatorname{Res}(f, ib) + \operatorname{Res}(f, ic)),$$ where $$f$$ is the integrand.

I addressed this question in a general form a couple weeks back! Here it is:

Let $$T$$ be a finite set of non zero real values. Then we may define the definite integral $$I(T)$$ as: $$$$I(T) = \int_{-\infty}^{\infty} \prod_{t \in T} \frac{1}{x^2 + t^2}\:dx \nonumber$$$$ To address this integral we decompose the integrand using a partial fraction decomposition: $$$$\prod_{t \in T} \frac{1}{x^2 + t^2} = \sum_{t \in T} \frac{A_t}{x^2 + t^2}\nonumber$$$$ Where $$A_t \in \mathbb{R}$$. Now: \begin{align} \prod_{t \in T} \frac{1}{x^2 + t^2} &= \sum_{t \in T} \frac{A_t}{x^2 + t^2} = \frac{A_{t_1}}{x^2 + t_1^2} + \frac{A_{t_2}}{x^2 + t_2^2} + \dots + \frac{A_{t_n}}{x^2 + t_n^2} \nonumber \end{align}

And so: $$$$1 = \sum_{t \in T} A_t\prod_{k \in T \setminus \{t\}} x^2 + k^2 \nonumber$$$$ Thus, $$$$A_t = \left[\prod_{k \in T \setminus \{t\}} k^2 - t^2\right]^{-1} \nonumber$$$$ We now can evaluate our integral $$I(T)$$: \begin{align} I(T) &= \int_{-\infty}^{\infty} \prod_{t \in T} \frac{1}{x^2 + t^2}\:dx = \sum_{t \in T} A_t \int_{-\infty}^{\infty}\frac{1}{x^2 + t^2}\:dx = \sum_{t \in T} A_t \left[\frac{1}{|t|}\arctan\left(\frac{x}{|t|}\right) \right]_{-\infty}^{\infty} \nonumber \\ &=\sum_{t \in T} A_t \cdot \frac{\pi}{|t|} = \pi \sum_{t \in T} \frac{1}{|t|} A_t \nonumber = \pi \sum_{t \in T} \left[|t| \prod_{k \in T \setminus \{t\}} k^2 - t^2\right]^{-1} \end{align}

And So we arrive at:

$$$$I(T) = \int_{-\infty}^{\infty} \prod_{t \in T} \frac{1}{x^2 + t^2}\:dx = \pi \sum_{t \in T} \frac{1}{|t|}\prod_{k \in T \setminus \{t\}}\frac{1}{k^2 - t^2} \nonumber$$$$

For your integral $$T = \left\{a,b,c \right\}$$