Complex Integration of $\int_0^\infty e^{-ax}\cos(bx)\,dx$

Question

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!

Answer 1

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 $V=\int_0^\infty e^{-ax}\sin(bx)\,dx$.

Answer 2

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 $R\to\infty$.

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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 $R\to\infty$.

Answer 3

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 $R \rightarrow \infty$.