I have been trying to evaluate this integral for a while now, $$I(\alpha,\beta,\gamma)= \int_{0}^{\infty} \cos\left(\frac{x(x^2-\alpha^2)}{x^2-\beta^2}\right)\frac{1}{x^2+\gamma^2} \, dx.$$ I first came across this integral on Quora (https://www.quora.com/What-improper-integrals-are-hard-to-solve).

Someone stated that this integral was submitted as a problem to the Gazette of the Royal Mathematics Society of Spain and was (at the time) still open, so the complete solution would not be published there. (have not been able to find the publication).

Although no derivation, a closed form expression was given: $$I(\alpha,\beta,\gamma)=\frac{\pi}{2\gamma}e^{-\frac{\gamma(\alpha^2+\gamma^2)}{\beta^2+\gamma^2}}.$$ One might be wondering, why not just respond to the post on Quora? That I did. However, the thread seems to inactive.

I have tried several methods including the one employed by Mark Viola in this post (Computing $\int_{-\infty}^{\infty} \frac{\cos x}{x^{2} + a^{2}}dx$ using residue calculus). Trying to work backwards from the closed form expression, by treating the term $ \frac{\gamma(\alpha^2+\gamma^2)}{\beta^2+\gamma^2}$ in exponent as one variable. However, I have not been able to succeed.

Hopefully, you can help me solve this integral/ give suggestions/ different methods.

edit: I found that when $\alpha=\beta$, the integral becomes $I= \int_{0}^{\infty} \cos(x)\frac{1}{x^2+\gamma^2}dx$ , which evaluates to $I=\frac{\pi}{2\gamma}e^{-\gamma}$. This is comparable to the desired result. The argument in the exponent is different however.

  • $\begingroup$ I can’t really work on this problem as of right now but have you tried Feynman’s Trick or rewriting $\cos x$ as the real part of $e^{ix}$? $\endgroup$
    – Frank W
    Sep 10, 2018 at 14:05
  • $\begingroup$ Have tried write the cosine in terms of exponentials followed by separating the integral and substitution of variables. As for Fyenman's trick, have attempted it, but tbh I wouldn't know with respect to what variable. $\endgroup$
    – Jameson
    Sep 10, 2018 at 14:36
  • $\begingroup$ What are the conditions on the parameters $\alpha, \beta$, and $\gamma$? Based on the solution it appears at least that $\gamma > 0$. $\endgroup$
    – kobe
    Sep 10, 2018 at 16:17
  • $\begingroup$ $\alpha, \beta$ are equal or greater than 0 and $\gamma$ is greater than 0 $\endgroup$
    – Jameson
    Sep 10, 2018 at 21:14
  • 3
    $\begingroup$ The originial problem can be found here. It was asked in the 66th edition of 'Problemas y Soluciones' as 'Problema 248'. The solution is given in the 73rd edition. Some subtleties regarding the application of the residue theorem are not discussed in great detail, however. $\endgroup$ Sep 10, 2018 at 22:01

1 Answer 1


For $\alpha, \beta \geq 0$ and $\gamma > 0$ let $$ f_{\alpha,\beta,\gamma} (z) = \frac{\exp \left(\mathrm{i} z \frac{z^2 - \alpha^2}{z^2 - \beta^2}\right)}{z^2 + \gamma^2} \, , \, z \in \mathbb{C} \setminus \{\pm\beta, \pm \mathrm{i} \gamma\} \, . $$ By symmetry we have $$ I (\alpha,\beta,\gamma) = \frac{1}{2} \int \limits_{-\infty}^\infty f_{\alpha,\beta,\gamma} (x) \, \mathrm{d} x \, .$$ Clearly, the integral exists for every combination of parameters and is bounded by $\frac{\pi}{2 \gamma}$ .

We can compute its value using the residue theorem. First note that we have $$ \operatorname{Res} (f_{\alpha,\beta,\gamma}, \mathrm{i} \gamma) = \frac{1}{2 \mathrm{i} \gamma} \exp \left(- \gamma \frac{\alpha^2 + \gamma^2}{\beta^2 + \gamma^2}\right) $$ and (by Jordan's lemma) $$ \lim_{R \to \infty} \int \limits_{\Gamma_R} f_{\alpha,\beta,\gamma} (z) \, \mathrm{d} z = 0 \, ,$$ where $\Gamma_R$ is a semi-circle of radius $R > 0$ in the upper half-plane. A naive application of the residue theorem therefore yields $$ I (\alpha,\beta,\gamma) = \frac{1}{2} \left[ 2 \pi \mathrm{i} \operatorname{Res} (f_{\alpha,\beta,\gamma}, \mathrm{i} \gamma) - \lim_{R \to \infty} \int \limits_{\Gamma_R} f_{\alpha,\beta,\gamma} (z) \, \mathrm{d} z \right]= \frac{\pi}{2 \gamma} \exp \left(- \gamma \frac{\alpha^2 + \gamma^2}{\beta^2 + \gamma^2}\right) \, .$$ While the result is indeed correct, this calculation is only valid for $\alpha = \beta$ . In this case we obtain (as already mentioned in the question) $I(\alpha,\alpha,\gamma) = \frac{\pi}{2 \gamma}\mathrm{e}^{- \gamma}$ .

If $\alpha \neq \beta$ , $f_{\alpha,\beta,\gamma}$ has essential singularities on the real axis, namely at $\pm \beta$ . Thus we need to deform our contour using small semi-circles $\gamma_\varepsilon (\pm \beta)$ of radius $\varepsilon > 0$ centred at these points and show that their contribution to the integral vanishes as $\varepsilon \to 0$ .

For simplicity, I will only discuss the case $\beta = 0$ in detail. We want to find $$ \int \limits_{\gamma_\varepsilon (0)} f_{\alpha,0,\gamma} (z) \, \mathrm{d} z = \varepsilon \int \limits_0^\pi \frac{\mathrm{e}^{\mathrm{i} \left[\phi + \varepsilon \mathrm{e}^{\mathrm{i}\phi} - \alpha^2 \varepsilon^{-1} \mathrm{e}^{-\mathrm{i} \phi}\right]}}{\gamma^2+\varepsilon^2 \mathrm{e}^{2 \mathrm{i} \phi}} \, \mathrm{d} \phi = \frac{\varepsilon}{\gamma^2} \int \limits_0^\pi \mathrm{e}^{\mathrm{i} \left[\phi - \alpha^2 \varepsilon^{-1} \mathrm{e}^{-\mathrm{i} \phi}\right]} \left[1 + \mathcal{O} (\varepsilon) \right] \, \mathrm{d} \phi \, .$$ The leading-order term can actually be calculated analytically: for $z \in \mathbb{C}$ we have $$ \int \limits_0^\pi \mathrm{e}^{\mathrm{i} \left[\phi - z \mathrm{e}^{-\mathrm{i} \phi}\right]} \, \mathrm{d} \phi = \mathrm{i} \left[(2 \operatorname{Si}(z) - \pi) z + 2 \cos(z)\right] \, . $$ The sine integral $\operatorname{Si}$ satisfies $\lim_{x \to \infty} \operatorname{Si}(x) = \frac{\pi}{2}$, so we obtain $$ \int \limits_{\gamma_\varepsilon (0)} f_{\alpha,0,\gamma} (z) \, \mathrm{d} z \sim \frac{\mathrm{i} \varepsilon}{\gamma^2} \left[(2 \operatorname{Si}(\alpha^2 \varepsilon^{-1}) - \pi) \alpha^2 \varepsilon^{-1} + 2 \cos(\alpha^2 \varepsilon^{-1})\right] \stackrel{\varepsilon \to 0}{\longrightarrow} 0 $$ as desired. Now we are allowed to apply the residue theorem , which yields \begin{align} I(\alpha,0,\gamma) &= \frac{1}{2} \left[ 2 \pi \mathrm{i} \operatorname{Res} (f_{\alpha,0,\gamma}, \mathrm{i} \gamma) - \lim_{R \to \infty} \int \limits_{\Gamma_R} f_{\alpha,0,\gamma} (z) \, \mathrm{d} z - \lim_{\varepsilon \to 0} \int \limits_{\gamma_\varepsilon (0)} f_{\alpha,0,\gamma} (z) \, \mathrm{d} z \right] \\ &= \frac{\pi}{2 \gamma} \exp \left(- \frac{\alpha^2 + \gamma^2}{\gamma}\right) \, . \end{align}

Almost the same calculation yields \begin{align} I(\alpha,\beta,\gamma) &= \frac{1}{2} \left[ 2 \pi \mathrm{i} \operatorname{Res} (f_{\alpha,\beta,\gamma}, \mathrm{i} \gamma) - \lim_{R \to \infty} \int \limits_{\Gamma_R} f_{\alpha,\beta,\gamma} (z) \, \mathrm{d} z - \lim_{\varepsilon \to 0} \int \limits_{\gamma_\varepsilon (\beta)} f_{\alpha,\beta,\gamma} (z) \, \mathrm{d} z - \lim_{\varepsilon \to 0} \int \limits_{\gamma_\varepsilon (-\beta)} f_{\alpha,\beta,\gamma} (z) \, \mathrm{d} z \right] \\ &= \frac{\pi}{2 \gamma} \exp \left(- \gamma \frac{\alpha^2 + \gamma^2}{\beta^2 + \gamma^2}\right) \end{align} for $\beta > 0$ .

Combining these results we conclude that the integral is given by $$ I(\alpha,\beta,\gamma) = \frac{\pi}{2 \gamma} \exp \left(- \gamma \frac{\alpha^2 + \gamma^2}{\beta^2 + \gamma^2}\right) $$ for every $\alpha,\beta \geq 0$ and $\gamma > 0$ .

  • 1
    $\begingroup$ It would seem that the OP seeks a real analysis only approach inasmuch as the cited reference uses contour integration. $\endgroup$
    – Mark Viola
    Sep 12, 2018 at 19:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.