3

$$\int_0^{\infty} {e^{-ax}\cos{(bx)}} dx$$

I know I need to use integration by part method. But I'm not sure how does the improper integral take place?

StationaryTraveller
  • 2,453

3 Answers3

5

Just an idea:

$$\cos bx=\frac{e^{ibx}+e^{-ibx}}{2}\implies \int\limits_0^\infty e^{-ax}\cos bx\,dx=\frac{1}{2}\int\limits_0^\infty\left(e^{-(a-ib)x}+e^{-(a+ib)x}\right)dx=\ldots$$

Check the values of $\,a,b\,$ (reals, I presume...?) for which the above converges.

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
  • to deal with improper integral just consider the following:

    $$\lim_{a \to +\infty} (\int_0^a f(x),dx) $$

    Integrate the expression inside the limit, and then, evaluate the limit itself.

     – Koba Apr 11 '13 at 14:24
5

I assume $a$ and $b$ are real numbers, and $a>0$.

Since $\cos{bx}=\operatorname{Re}\{e^{-ibx}\}$, we get $$ \int_0^\infty e^{-ax}\cos(bx)\,dx= \operatorname{Re}\int_0^\infty e^{-(a+ib)x}\,dx= \operatorname{Re}\left[-\frac{e^{-(a+ib)x}}{a+ib}\right]_0^\infty= $$ $$ \operatorname{Re}\left\{\frac{1}{a+ib}\right\}= \operatorname{Re}\left\{\frac{a-ib}{a^2+b^2}\right\}= \frac{a}{a^2+b^2}. $$

Mårten W
  • 3,480
3

We can easily get this

$$\int \frac{\cos(bx)dx}{e^{ax}}=\frac{b\sin(bx)-a\cos(bx)}{(a^2+b^2)e^{ax}}+C$$

and you only need to get the limit to the infinity

$$\frac{b\sin(bx)-a\cos(bx)}{(a^2+b^2)e^{ax}}=\frac{\sqrt{a^2+b^2}\sin(bx+\phi)}{(a^2+b^2)e^{ax}}<\frac{\sqrt{a^2+b^2}}{(a^2+b^2)e^{ax}}$$

so

$$\lim_{x\rightarrow+\infty}\frac{\sqrt{a^2+b^2}\sin(bx+\phi)}{(a^2+b^2)e^{ax}}=0$$

Xiaolang
  • 495
  • 1
    Integration by parts. Yes. Not sure why this was downvoted, as this seems to be the only answer which is actually answering the question :-) Perhaps you need to show more details of how you got that indefinite integral, Xiaolang. – Aryabhata Apr 11 '13 at 14:32
  • Yes Idont know why it was downvoted too,and i calculate the limit but i'm not sure it was actually true ,you can check it. – Xiaolang Apr 11 '13 at 14:34
  • @Aryabhata it is the only elementary solution. the other ones do make sense but appeals to complex integration – Lost1 Apr 11 '13 at 14:49
  • @Lost1 yes,because i'm a new student in the university and i can't use the knowledge in complex anylasis,if OP wants a advanced solution I can just say sorry :-) – Xiaolang Apr 11 '13 at 14:51
  • @Lost1: Yes, and integration by parts was specifically mentioned in the question. Not sure why OP chose a different answer. – Aryabhata Apr 11 '13 at 14:51
  • @Xiaolang: You probably should elaborate on how you actually used integration by parts to help the question asker better understand. Right now, the first line might seem like a rabbit out of a hat. – Aryabhata Apr 11 '13 at 14:52
  • i think the method is very (i dont know how to describe this feeling) let $I_1=\int e^{ax}\sin(bx)dx $ and $I_2=\int e^{ax}\cos(bx)dx$ and use the integration by parts we can get a obvious relationship of $I_1$ and $I_2$ – Xiaolang Apr 11 '13 at 14:55