Its been awhile since I have taken complex analysis and I am wondering how to solve the following integral when $a>0, \ a=0,$ and $a<0$ for $$\int^{\infty}_0\frac{\cos ax+x\sin ax}{1+x^2}dx.$$

I know I can use the residue theorem thus $$\int^{\infty}_0\left(\frac{\cos ax}{1+x^2}+\frac{x\sin ax}{1+x^2}\right)dx.$$

  • $\begingroup$ The usual trick is using $\exp(az)/(1+z^2)$ and $z\exp(az)/(1+z^2)$. $\endgroup$ Dec 1, 2014 at 8:57
  • $\begingroup$ Somewhat a related problem by Ron Gordon $\endgroup$
    – Venus
    Dec 1, 2014 at 10:11
  • $\begingroup$ @Martín-BlasPérezPinilla You mean $\exp(i\,a\,z)$ don't you? $\endgroup$ Dec 1, 2014 at 10:49
  • $\begingroup$ @JuliánAguirre, obviously. $\endgroup$ Dec 1, 2014 at 11:23
  • $\begingroup$ Note that $~\displaystyle\int_0^\infty\frac{x\sin ax}{1+x^2}dx=-\frac d{da}\int_0^\infty\frac{\cos ax}{1+x^2}dx$. $\endgroup$
    – Lucian
    Dec 1, 2014 at 14:13

3 Answers 3


First note that the integral may be written as

$$\frac12 \int_{-\infty}^{\infty} dx \frac{\cos{a x}}{1+x^2} - \frac12 \frac{d}{da} \int_{-\infty}^{\infty} dx \frac{\cos{a x}}{1+x^2} $$

So consider the contour integral

$$\oint_{C(a)} dz \frac{e^{i a z}}{1+z^2} $$

where $C(a)$ is a semicircle of radius $R$ in the upper half plane when $a\gt 0$ and in the lower half plane when $a \lt 0$. For example, when $a \gt 0$, the integral is

$$\int_{-R}^R dx \frac{e^{i a x}}{1+x^2} + i R \int_0^{\pi} d\phi \, e^{i \phi} \frac{e^{i a R \cos{\phi}} e^{-a R \sin{\phi}}}{1+R^2 e^{i 2 \phi}} $$

As $R \to \infty$, the magnitude of the second integral is bounded by

$$\frac{2 R}{R^2-1} \int_0^{\pi/2} d\phi \, e^{-2 a R \phi/\pi} = \frac{\pi}{a (R^2-1)} \left ( 1-e^{-a R} \right )$$

which clearly vanishes in this limit. On the other hand, the contour integral is also equal to $i 2 \pi$ times the residue at the pole $z=i$, so that

$$\int_{-\infty}^{\infty} dx \frac{\cos{a x}}{1+x^2} = i 2 \pi \frac{e^{-a}}{2 i} = \pi \, e^{-a} \quad (a\gt 0)$$

(Here we used the symmetry of the integrand to get the original integral back.) For $a \lt 0$, we close in the lower half plane (or simply use the evenness of the cosine) and find that

$$\int_{-\infty}^{\infty} dx \frac{\cos{a x}}{1+x^2} = \pi \, e^{-|a|} $$

We then take the derivative of this integral (separately for the cases $a \gt 0$ and $a \lt 0$) find that the original integral is

$$\int_0^{\infty} dx \frac{\cos{a x} + x \sin{a x}}{1+x^2} = \frac{\pi}{2} \left (1+\operatorname{sgn}{a} \right )e^{-|a|} = \pi \theta(a) e^{-|a|}$$

where $\theta$ is the Heaviside step function, which is equal to $1/2$ when $a=0$.

  • $\begingroup$ Thank you very much!! How did you get $e^{iaz}$ from $\cos ax$? $\endgroup$
    – Robben
    Dec 1, 2014 at 20:28
  • 1
    $\begingroup$ @Robben: I used symmetry to replace the integration inetrval to a symmetric interval. This then allowed me to substitute the cosine with the exponential, because the sine piece then contributes nothing to the integral. $\endgroup$
    – Ron Gordon
    Dec 1, 2014 at 20:37

Alternatively, you could consider $ \displaystyle f(z) = \frac{e^{iaz}}{z-i}$.

Then for $a >0$ we can use Jordan's lemma to conclude $$\int_{-\infty}^{\infty} \frac{e^{iax}}{x-i} \ dx = 2 \pi i \ \text{Res}[f(z),i] = 2 \pi i e^{-a}. $$

But $$ \begin{align} \int_{-\infty}^{\infty} \frac{e^{iax}}{x-i} \ dx &= \int_{-\infty}^{\infty} \frac{\Big(\cos(ax) + i \sin(ax) \Big)(x+i)}{x^{2}+1} \ dx \\ &= \int_{-\infty}^{\infty} \frac{x \cos (ax) - \sin(ax) + i \Big(\cos(ax) + x \sin(ax) \Big)}{x^{2}+1} \ dx . \end{align}$$

And equating the imaginary parts on both sides of the equation we get $$\int_{0}^{\infty} \frac{\cos (ax) + x \sin(ax)}{x^{2}+1} \ dx = \pi e^{-a}. $$

Now if $a<0$, $ \displaystyle \int f(z) \ dz$ will vanish along the lower half of the circle $|z|=R$ as $R \to \infty$.

And since $f(z)$ is analytic in the lower half-plane, the integral in question evaluates to zero.

  • $\begingroup$ @M.N.C.E. Thanks. $\endgroup$ Dec 4, 2014 at 7:20
  • $\begingroup$ Hi, I want to ask unrelated question. Where did you encounter this problem $$\int_0^\frac{1}{\sqrt 2} \frac{\sqrt{1+x^4}}{1-x^4} dx = \frac{1}{2 \sqrt 2}\left[\frac \pi 2 + \ln(2 + \sqrt 5) - \arctan \frac{\sqrt 5}{2}\right]$$Can we discuss this in chat room? Thanks. $\endgroup$
    – Venus
    Dec 4, 2014 at 11:07

First of all for $a=0$ you have $$\int^{\infty}_{0}\Big(\frac{\cos(ax)}{1+x^2}+\frac{x\sin(ax)}{1+x^2}\Big)\,dx=\int^{\infty}_{0}\frac{1}{1+x^2}\,dx=\arctan(x)\Big|^{\infty}_0=\frac{\pi}{2}$$

In general for $a\neq 0$ we use
$$\cos(ax)=\frac{e^{iax}+e^{-iax}}{2}$$ and $$\sin(ax)=\frac{e^{iax}-e^{-iax}}{2i}$$ hence your integral can be rewritten as $$\int^{\infty}_{0}\Big(\frac{\cos(ax)}{1+x^2}+\frac{x\sin(ax)}{1+x^2}\Big)\,dx=\int^{\infty}_{0}\Big(\frac{e^{iax}+e^{-iax}}{2(1+x^2)}+\frac{x(e^{iax}-e^{-iax})}{2i(1+x^2)} \Big)\,dx$$ Substitute $z=ix$ to get $$\int^{i\infty}_{0}\Big(\frac{e^{az}+e^{-az}}{2(1-z^2)}-\frac{z(e^{az}-e^{-az})}{2(1-z^2)} \Big)\frac{dz}{i}=\frac{1}{4}\int^{i\infty}_{-i\infty}\Big(\frac{e^{az}+e^{-az}}{1-z^2}-\frac{z(e^{az}-e^{-az})}{1-z^2} \Big)\frac{dz}{i}$$ Let $$\mathcal{I}=\frac{1}{4}\int^{i\infty}_{-i\infty}\Big(\frac{e^{az}+e^{-az}}{1-z^2}-\frac{z(e^{az}-e^{-az})}{1-z^2} \Big)\frac{dz}{i}=\frac{1}{4}\int^{i\infty}_{-i\infty}\Big(\frac{(1-z)+(1+z)e^{-2az}}{1-z^2}\Big)\cdot e^{az}\frac{dz}{i}$$ The last integral can be partitioned into two integrals as follows $$\mathcal{I}=\int^{i\infty}_{-i\infty}\frac{1}{4(1+z)}\cdot e^{az}\frac{dz}{i}+\int^{i\infty}_{-i\infty}\frac{1}{4(1-z)}\cdot e^{-az}\frac{dz}{i}$$ Substituting $z\equiv-z$ for the second integral yields $$\mathcal{I}=\int^{i\infty}_{-i\infty}\frac{1}{4(1+z)}\cdot e^{az}\frac{dz}{i}-\int^{-i\infty}_{i\infty}\frac{1}{4(1+z)}\cdot e^{az}\frac{dz}{i}=\frac{1}{2i}\int^{i\infty}_{-i\infty}\frac{1}{1+z}\cdot e^{az}\,dz$$ Now we will do an analysis of how to evaluate the very last integral. Firstly consider the following contour integral $$\mathcal{J}=\oint_{\gamma}\frac{e^{az}}{1+z}\,dz$$ where $\Re(a)>0$ and $\gamma$ is the contour which consists of the imaginary axis and half disk on the left-half complex plane. The orientation we choose to be counter-clockwise. We can now partition the contour integral into its component parts, one along the imaginary axis and the other along the half disk on the left-half plane keeping the chosen orientation, i.e. $$\oint_{\gamma}\frac{e^{az}}{1+z}\,dz=\oint_{\Gamma_{R}}\frac{e^{az}}{1+z}\,dz+\int^{+iR}_{-iR}\frac{e^{az}}{1+z}\,dz$$ where the first part is along the half disk and the other along the imaginary axis with end points $-iR$ and $+iR$. The integral along the half disk can be estimated as follows by using the ML estimate i.e. $$\Big|\oint_{\Gamma_{R}}\frac{e^{az}}{1+z}\,dz\Big|=\Big|\int^{3\pi/2}_{\pi/2}\frac{e^{aRe^{i\theta}}}{1+Re^{i\theta}}Rie^{i\theta}\,d\theta\Big|\leq\int^{3\pi/2}_{\pi/2}\frac{|e^{aRe^{i\theta}}|}{|1+Re^{i\theta}|}|Rie^{i\theta}||\,d\theta|\leq \frac{e^{-aR}}{R-1}R\cdot \pi\sim e^{-aR}\to 0$$ as $R\to\infty$. Thus in the limit the only contribution comes from the second integral i.e. $$\lim_{R\to\infty}\oint_{\gamma}\frac{e^{az}}{1+z}\,dz=\lim_{R\to\infty}\Big(\oint_{\Gamma_{R}}\frac{e^{az}}{1+z}\,dz+\int^{+iR}_{-iR}\frac{e^{az}}{1+z}\,dz\Big)$$$$=\lim_{R\to\infty}\oint_{\Gamma_{R}}\frac{e^{az}}{1+z}\,dz+\lim_{R\to\infty}\int^{+iR}_{-iR}\frac{e^{az}}{1+z}\,dz=\lim_{R\to\infty}\int^{+iR}_{-iR}\frac{e^{az}}{1+z}\,dz$$ On the other hand the integrand of the contour integral is analytic on the domain defined by its contour except for a simple pole at $z=-1$ with residue $e^{-a}$. Appealing to Cauchy Residue Theorem we obtain $$\mathcal{J}=\oint_{\gamma}\frac{e^{az}}{1+z}\,dz=2\pi i\cdot\text{Res}\{\frac{e^{az}}{1+z},-1\}=2\pi i e^{-a}$$ As $R\to\infty$ the analytic properties of the integrand don't change so the residue is the same. Hence $$\lim_{R\to\infty}\int^{+iR}_{-iR}\frac{e^{az}}{1+z}\,dz=2\pi i e^{-a}$$ (This is related to Laplace transform of $\frac{1}{1+z}$ with transform variable $a$ such that $\Re{(a)}>0$
$$\mathcal{L}^{-1}(\frac{1}{1+z})(a)=\lim_{T\to\infty}\frac{1}{2\pi i}\int^{\sigma+iT}_{\sigma-iT}\frac{1}{1+z}\cdot e^{az}\,{dz}=\frac{1}{2\pi i}\int^{\sigma+i\infty}_{\sigma-i\infty}\frac{1}{1+z}\cdot e^{az}\,{dz}=e^{-a}$$ with $\sigma=0$) Having evaluated $\mathcal{J}$ we can now find $\mathcal{I}$ as $$\mathcal{I}=\frac{1}{2i}\mathcal{J}=\pi e^{-a}$$ For $\Re(a)<0$ we set $a=-b$ with $\Re(b)>0$ and $\mathcal{I}$ can be rewritten as $$\mathcal{I}=\frac{1}{2i}\int^{i\infty}_{-i\infty}\frac{1}{1+z}\cdot e^{az}\,dz=\frac{1}{2i}\int^{i\infty}_{-i\infty}\frac{1}{1-z}\cdot e^{bz}\,dz$$ We have used in the last step the substitution $z\equiv-z$. Consider $$\mathcal{J}=\oint_{\gamma}\frac{e^{bz}}{1-z}\,dz$$ where $\gamma$ is the same contour as above. However the integrand is analytic everywhere on the domain defined by the contour so Cauchy Residue Theorem will tell us that $$\mathcal{J}=\oint_{\gamma}\frac{e^{bz}}{1-z}\,dz=0$$ As the radius of the half disk increases i.e. $R\to\infty$ the analytic properties of the integrand don't change so the contour integral evaluates to zero. But in the limit you can show by the same arguments as above (ML estimate) that the only contribution if there would be any would come from the integral on the imaginary axis i.e. $$\lim_{R\to\infty}\oint_{\gamma}\frac{e^{bz}}{1-z}\,dz=\lim_{R\to\infty}\int^{+iR}_{-iR}\frac{e^{bz}}{1-z}\,dz$$ Together with the result from Cauchy we obtain $$\mathcal{J}=\lim_{R\to\infty}\int^{+iR}_{-iR}\frac{e^{bz}}{1-z}\,dz=0$$ On the other hand $$\mathcal{I}=\frac{1}{2i}\mathcal{J}=0$$ So in conclusion we have $$ \mathcal{I} = \left\{ \begin{array}{ll} 0 & \quad \Re(a)<0 \\ \frac{\pi}{2} & a=0\\ \pi e^{-a} & \Re(a)>0 \end{array} \right. $$

  • $\begingroup$ You made a mistake somewhere; your end result is not correct. $\endgroup$
    – Ron Gordon
    Dec 1, 2014 at 12:30
  • $\begingroup$ it should be ok now I think. $\endgroup$
    – Arian
    Dec 1, 2014 at 12:37
  • $\begingroup$ Nope. The result should be zero for $a \lt 0$. $\endgroup$
    – Ron Gordon
    Dec 1, 2014 at 12:38
  • $\begingroup$ Something tells me that your mistake lies in the fact that, in order to use residues for the inverse Laplace transform, you need to understand the integration contour you are using. You've done none of that analysis. $\endgroup$
    – Ron Gordon
    Dec 1, 2014 at 12:44
  • 1
    $\begingroup$ This is much improved over the previous "solution." That said, you also somewhat complicated the case $a \lt 0$. All you need to do is close the contour to the right (because that is where the integral about the semicircle vanishes). Then you'll see that the integral is zero by the residue theorem, etc. $\endgroup$
    – Ron Gordon
    Dec 2, 2014 at 13:49

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