How to compute $\int_0^\infty e^{-ax}cos(bx) dx $?

Question

I need to compute $\int_0^\infty e^{-ax}cos(bx) dx $. Integration by parts seems to go nowhere and I can't think of any useful substitutions.

See here: http://math.stackexchange.com/questions/358356/solving-the-improper-integral-int-0-infty-e-ax-cosbx-dx?rq=1 — Brenton, Mar 18 '16 at 19:18
Integration by parts works fine if you call the integral $I$, do by parts twice and rearrange to get $I$ which reappears — David Quinn, Mar 18 '16 at 19:19

score 4 · Answer 1 · answered Mar 18 '16 at 19:22

4

We can use $\cos(bx)=\Re(e^{ibx})$ to obtain

\begin{eqnarray} I\equiv \Re \int_0^\infty e^{-ax} e^{ibx}dx=\Re \int_0^\infty e^{-\alpha x} dx=\Re\left(\frac{1}{\alpha}\right) \end{eqnarray} where $\alpha \equiv a-ib$, assuming $a>0$. Thus we obtain $$ I= \Re\left(\frac{1}{\alpha}\right)=\Re\left( \frac{1}{a-ib}\right)=\Re\left( \frac{a+ib}{a^2+b^2}\right) $$ You can see that \begin{equation} I= \frac{a}{a^2+b^2} \end{equation}

answered Mar 18 '16 at 19:22

Jeff Faraci

9,878

Can you tell me why $\int_{0}^{\infty}e^{-\alpha x}dx=\frac{1}{\alpha}$ where $\alpha$ is a complex number $x$ is real!? – Hosein Rahnama Apr 12 '18 at 14:33

score 1 · Accepted Answer · answered Mar 18 '16 at 19:25

Assume $\;a>0\;$ and put

$$I=\int_0^\infty e^{-ax}\cos bx\,dx$$

Now, by parts:

$$\begin{cases}u=\cos bx\implies&u'=-b\sin bx\\{}\\v'=e^{-ax}\implies&v=-\frac1ae^{-ax} \end{cases}\;\;\implies I=\left.-\frac{e^{-ax}\cos bx}a\right|_0^\infty-\frac ba\int_0^\infty e^{-ax}\sin bx\,dx$$

and again by parts and take the limits:

$$I=\frac1a+\left.\frac b{a^2} e^{-ax}\sin bx\right|_0^\infty-\frac{b^2}{a^2}I\implies\left(\frac{a^2+b^2}{a^2}\right)I=\frac1a\implies I=\frac a{a^2+b^2}$$

How to compute $\int_0^\infty e^{-ax}cos(bx) dx $?

2 Answers2