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Out of Stein's book, we're asked to show find a formula for $$\int_0^\infty e^{-ax}\cos(bx)\,dx,\quad a>0.$$While this is very doable via integration by parts, I'm asked to use contour integration, where we're suggested to integrate over a sector with angle $\omega$ such that $\cos(\omega)=a/\sqrt{a^2+b^2}.$

I've attempted this multiple times, and I keep having trouble with some integrals. I've set the contour up so that on the first segment, it's on the real axis, so we have the integral $$\int_0^R e^{-az}\cos(bz)\,dz.$$ Then I parameterize the arc as $z(\theta)=Re^{i\theta}$ for $0\leq\theta\leq \omega$, so the second integral becomes $$\int_0^\omega e^{-a(Re^{i\theta})}\cos(b(Re^{i\theta}))\left(iRe^{i\theta}\right)\,d\theta.$$The final segment I parameterized as $z(t)=Re^{i\omega}(1-t)$ and set up the final integral as $$\int_0^1e^{-a(Re^{i\omega}(1-t))}\cos\big(b(Re^{i\omega}(1-t))\big)(-Re^{i\omega})\,dt.$$I've tried finding some way to bound one of the last two integrals so that I can show one of them goes to $0$ as $R\to\infty$, but I've not had any luck. Will someone make a suggestion if my approach and parameterizations are correct? Thanks!

Update: My thoughts are really that the integral which goes to zero is the arc. I keep working it down in the following way; we know that it is \begin{align} &\leq R\int_0^\omega\left|e^{-aR(\cos\theta+i\sin\theta)}\cdot\left(\frac{e^{ibRe^{i\theta}}+e^{-ibRe^{i\theta}}}{2}\right)\right|\,d\theta\\ &\leq\frac{R}{2}\int_0^\omega\left|e^{-aR\cos\theta}\cdot\left(e^{ibR(\cos\theta+i\sin\theta)}+e^{-bR(\cos\theta+i\sin\theta)}\right)\right|\,d\theta\\ &\leq\frac{R}{2}\int_0^\omega\left|e^{-aR\cos\theta-bR\sin\theta}\right|+\left|e^{-aR\cos\theta+bR\sin\theta}\right|\,d\theta. \end{align} At this point, it is easy to show that the first term tends to zero, since $(-aR\cos\theta)<0$ and $bR\sin\theta>0$ (since $b$ and $\sin\theta$ have the same sign). The second term, however is what causes me trouble. I just finished working it out again, and I get that it only goes to zero if $a^2>b^2$, which isn't necessary in the general formula when achieved by integration by parts. I am really at a loss...

Added Solution: See the solution I've posted and please leave comments on your thoughts about it. Thanks!

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  • $\begingroup$ Is this the exercise? books.google.com/books/… $\endgroup$
    – Erick Wong
    Feb 11, 2013 at 7:30
  • $\begingroup$ +1 I have never seen like this one just in Riemannian integral for reals. $\endgroup$
    – Mikasa
    Feb 17, 2013 at 19:46

3 Answers 3

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We have \begin{align*} &\int_0^Re^{-Az}\,dz+\int_0^{\omega}e^{-A(Re^{i\theta})}iRe^{i\theta}\,d\theta+\int_0^Re^{-A(R-t)e^{i\omega}}(-e^{i\omega})\,dt\\ &=\frac{1}{A}+\underbrace{\int_0^{\omega}e^{-A(Re^{i\theta})}iRe^{i\theta}\,d\theta}_{\text{vanishes as $R\longrightarrow\infty$}}-\int_0^Re^{-At(\cos\omega+i\sin\omega)}(e^{i\omega})\,dt\\ &=\frac{1}{A}-\int_0^Re^{-at-ibt}\left(\frac{a}{A}+i\frac{b}{A}\right)\,dt\\ &=\frac{1}{A}-\frac{a}{A}\int_0^Re^{-at-ibt}\,dt-i\frac{b}{A}\int_0^Re^{-at-ibt}\,dt\\ &=\frac{1}{A}-\frac{a}{A}\int_0^Re^{-at}\big(\cos(bt)-i\sin(bt)\big)\,dt-i\frac{b}{A}\int_0^Re^{-at}\big(\cos(bt)-i\sin(bt)\big)\,dt\\ &=\frac{1}{A}-\frac{a}{A}\int_0^Re^{-at}\cos(bt)\,dt-\frac{b}{A}\int_0^Re^{-at}\sin(bt)\,dt\\ &\hspace{0.5in}+i\left(\frac{a}{A}\int_0^Re^{-at}\sin(bt)\,dt-\frac{b}{A}\int_0^{R}e^{-at}\cos(bt)\,dt\right). \end{align*}

Argument for Vanishing Here the integral over the arc vanishes since $$\begin{align*} \left|\int_0^{\omega}e^{-A(Re^{i\theta})}iRe^{i\theta}\,d\theta\right|&\leq R\int_0^\omega\left|e^{-AR\cos\theta}\right|\,d\theta\\&\leq R\int_0^\omega e^{-AR(1-2\theta/\pi)}\,d\theta \end{align*}$$and now it is easy to show that the integral goes to $0$.

Finsihed Solution Now we can set this up as a linear system since we know the real part and imaginary parts must be zero; so we have the equations \begin{align*} &\frac{1}{A}-\frac{a}{A}U-\frac{b}{A}V=0\\ &\hspace{0.15in}-\frac{b}{A}U+\frac{a}{A}V=0. \end{align*} Solving the system yields $$ U=\frac{a}{a^2+b^2} $$ and $$ V=\frac{b}{a^2+b^2} $$ as desired, where $U=\int_0^\infty e^{-ax}\cos(bx)\,dx$ and $V=\int_0^\infty e^{-ax}\sin(bx)\,dx$.

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  • $\begingroup$ I really like this answer, but I am wondering what happened to the R term on the third integral in the first equation: It goes $$\int_0^Re^{-A(R-t)e^{iw}}(-e^{iw})=-\int_0^Re^{-Ate^{iw}}(e^{iw})$$ Can you elaborate on what happened here? Thanks. $\endgroup$
    – Mike
    Apr 9, 2016 at 6:26
  • $\begingroup$ @Craig: Just use substitution with $u=R-t$. $\endgroup$
    – Clayton
    Apr 9, 2016 at 21:54
  • $\begingroup$ I see, in which case the upper bound of the integral would be $u+t$, and safely you might as well call this $R$ because it's tending to infinity all the same? $\endgroup$
    – Mike
    Apr 9, 2016 at 22:01
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    $\begingroup$ @Craig: Not quite. The upper bound of the integral would be $0$, the lower bound is $R$, and we get a negative factor from the fact that $du=-dt$. This allows you to switch the bounds of the integral without problem. $\endgroup$
    – Clayton
    Apr 13, 2016 at 16:45
  • $\begingroup$ I have two questions @Clayton. The first integral on the contour should be $\int_{0}^{R} e^{-Ax} dx$ since we are traversing only on the x-axis. Also, in the third integral why do you have $R-t$ ? I took it as $\int_{0}^{R} e^{-Ate^{iw} }e^{iw} dt$ since it is over the line with angle $iw$ going from 0 to R. $\endgroup$
    – u_any_45
    Sep 20, 2020 at 4:05
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As long as $a>0$, $$ \begin{align} \int_0^\infty e^{-ax}\cos(bx)\,\mathrm{d}x &=\mathrm{Re}\left(\int_0^\infty e^{-(a-ib)x}\mathrm{d}x\right)\\ &=\mathrm{Re}\left(\int_0^\infty e^{-(a^2+b^2)x}\mathrm{d}(a+ib)x\right)\tag{$\ast$}\\ &=\mathrm{Re}\left(\frac{a+ib}{a^2+b^2}\int_0^\infty e^{-(a^2+b^2)x}\mathrm{d}(a^2+b^2)x\right)\\ &=\mathrm{Re}\left(\frac{a+ib}{a^2+b^2}\right)\\ &=\frac{a}{a^2+b^2} \end{align} $$ Step $(\ast)$ is simply adding $$ \mathrm{Re}\left(\int_\gamma e^{-(a-ib)z}\,\mathrm{d}z\right) $$ where $\gamma$ is the contour that goes from $0$ to $|z|=R$ along the curve $(a+ib)x$ ($x$ from $0$ to $\infty$), then follows $|z|=R$ to the positive real axis, and then back along the real axis to $0$. There are no singularities inside $\gamma$ so the integral is $0$. The integral along the arc of $|z|=R$ vanishes as $R\to\infty$.


Why the integral along the curve vanishes

Without loss of generality, assume $b>0$. Parameterizing $\gamma(t)=R(\cos(t)+i\sin(t))$ and setting $\tan(\theta)=b/a$ $$ \begin{align} &\left|\,\int_0^\theta e^{-(a-ib)R(\cos(t)+i\sin(t))}\,\mathrm{d}R(\cos(t)+i\sin(t))\,\right|\\ &=\left|\,R\int_0^\theta e^{-\sqrt{a^2+b^2}R(\cos(t-\theta)+i\sin(t-\theta))}\,i(\cos(t)+i\sin(t))\,\mathrm{d}t\,\right|\\ &\le R\int_0^\theta e^{-\sqrt{a^2+b^2}R\cos(t-\theta)}\,\mathrm{d}t\\ &=R\int_0^\theta e^{-\sqrt{a^2+b^2}R\cos(t)}\,\mathrm{d}t\\ &\le R\int_0^\theta e^{-\sqrt{a^2+b^2}R(1-2t/\pi)}\,\mathrm{d}t\\ &=Re^{-\sqrt{a^2+b^2}R}\int_0^\theta e^{\sqrt{a^2+b^2}R(2t/\pi)}\,\mathrm{d}t\\ &=\frac{\pi/2}{\sqrt{a^2+b^2}} Re^{-\sqrt{a^2+b^2}R}\left(e^{\sqrt{a^2+b^2}R(2\theta/\pi)}-1\right)\\ &=\frac{\pi/2}{\sqrt{a^2+b^2}} R\left(e^{\sqrt{a^2+b^2}R(2\theta/\pi-1)}-e^{-\sqrt{a^2+b^2}R}\right)\tag{$\lozenge$} \end{align} $$ Since $0\le\theta\lt\pi/2$, the coefficient of $R$ in each exponential is negative. Thus, $(\lozenge)$ tends to $0$ as $R\to\infty$.

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  • $\begingroup$ I think that last statement is what the OP wanted to prove. I made an attempt in my answer. I really like how the integrals only converge over the arc and nowhere else. $\endgroup$
    – Ron Gordon
    Feb 11, 2013 at 17:32
  • $\begingroup$ @rlgordonma: I have given a detailed proof of the vanishing of the integral along $|z|=R$. I hope it is not too obscure. $\endgroup$
    – robjohn
    Feb 11, 2013 at 18:29
  • $\begingroup$ I get it. It feels less "natural" than mine, but from my quick read, I see that it works because $\cos{t} \ge 1 - (2/\pi) t$ over $t \in (0,\pi/2)$. Bravo. (+1) $\endgroup$
    – Ron Gordon
    Feb 11, 2013 at 19:19
  • $\begingroup$ @robjohn: This seems like a slick proof, but the ($*$) and immediately thereafter loses me just a little bit. Want to elaborate that point a little bit? By the way, the explanation why the integral vanished on the arc is superb. +1$ $\endgroup$
    – Clayton
    Feb 12, 2013 at 0:49
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I think your approach is OK, at least technically correct, but why keep the integration limits of the wedge boundary at $(0,1)$, when you want them to be $(0,\infty)$? Viz,

$$\oint_C dz \: e^{-a z} e^{i b z} = 0$$

by Cauchy's Theorem. So, yes $C$ has 3 pieces, one being the pos. real line, one an arc, and the other the line $z=e^{i \omega}t$:

$$ e^{i \omega} \int_{-\infty}^0 dt \: \exp{[-(a-i b) (\cos{\omega} + i \sin{\omega}) t]} + \int_0^{\infty} dx \: e^{-a x} e^{i b x} + \int_{C_R} dz \: e^{-a z} e^{i b z} = 0 $$

where $C_R$ is the arc of the wedge. I will show below that the integral over $C_R$ vanishes. That leaves the other two integrals, which ad to zero. Rearranging the sum, we get

$$\int_0^{\infty} dx \: e^{-a x} e^{i b x} = e^{i \omega} \int_0^{\infty} dt \: e^{-\sqrt{a^2+b^2} t}$$

Taking the real part of both sides:

$$\int_0^{\infty} dx \: e^{-a x} \cos{(b x)} = \frac{a}{a^2+b^2}$$

Going back to the integral over the arc, we find it takes the form, depending on the sign of $b$,

$$\int_{C_R} dz \: e^{-a z} e^{\pm i b z} = i R \int_0^{\omega} d \theta \: e^{i \theta} \exp{[-a R (\cos{\theta} + i \sin{\theta})]} \exp{[\pm i b R (\cos{\theta} + i \sin{\theta})]} $$

which is bounded by

$$R \int_0^{\omega} d \theta \: \exp{[-R (a \cos{\theta} - b \sin{\theta})]} $$

Note that the argument of the exponent is positive only within the integration range, which works out perfectly. Therefore, on the arc of the wedge, the integral vanishes in the limit as $R \rightarrow \infty$.

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  • $\begingroup$ It is probably not too difficult, but you should probably show that $a\cos(\theta)-b\sin(\theta)\ge c\gt0$ so that the coefficient of $R$ is $\le-c$. $\endgroup$
    – robjohn
    Feb 11, 2013 at 19:40
  • $\begingroup$ @robjohn and rlgordonma: I don't see why it should be the case that $a\cos\theta-b\sin\theta>0$ at all, since this is equivalently $a\cos\theta>b\sin\theta$, thus $1>(b/a)\tan\theta$. The maximum tangent is in this range is $b/a$, thus we end up with $a^2>b^2$, which is what I got at the end... Help? $\endgroup$
    – Clayton
    Feb 12, 2013 at 0:23
  • $\begingroup$ @Clayton: between $0$ and $\omega$, $a \cos{\theta} - b \sin{theta}$ goes between $a$ and $0$. $\endgroup$
    – Ron Gordon
    Feb 12, 2013 at 1:37
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    $\begingroup$ @Clayton We would have $\tan \theta \le a/b$ rather than $b/a$, hence $a \cos \theta \ge b \sin \theta$. $\endgroup$
    – Erick Wong
    Feb 12, 2013 at 2:25
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    $\begingroup$ @Clayton: you seem to be thinking about the fact that in the posed problem, the sign of $b$ does not matter since $\cos(bx)=\cos(-bx)$. However, in carrying out the contour integration, we need to make sure that the integral along the arc of the wedge vanishes, and for this to occur, we need to match the sign of $b$ and the particular character, $e^{-(a+ib)x}$ or $e^{-(a-ib)x}$, we are using with the contour we are using. $\endgroup$
    – robjohn
    Feb 12, 2013 at 12:22

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