Using integration by parts show that:

Question

$$\int_0^{+\infty} e^{-x} \cos x\, dx = \int_0^{+\infty} e^{-x} \sin x\, dx$$

So far I have just expanded out, and I don't see an end, is there a trick I am missing?

Answer 1

A rather simple and detailed approach :

$$\int_0^\infty e^{-x}\cos(x)\mathrm{d}x = \int_0^\infty e^{-x}(\sin(x))'\mathrm{d}x = \big[ e^{-x}\sin(x)\big]_0^\infty - \int_0^\infty(e^{-x})'\sin(x)\mathrm{d}x$$

$$=$$

$$\lim_{x\to \infty} \big[ e^{-x}\sin(x)\big] - 0 - \int_0^\infty-e^{-x}\sin(x)\mathrm{d}x $$

$$=$$

$$\boxed{\int_0^\infty e^{-x}\sin(x)\mathrm{d}x}$$

Answer 2

apply integration by part twice then you'll get the same integral with extra terms then call integral I then it is a first order linear equation so, solve it for I. Other approach add $$\int e^{-x} i \cos(x)$$ to original integral. So desired result is real part of $$\int_{0}^{\infty} e^{-x}(\sin(x)+ i \cos(x))dx = \int e^{(-1+i)x} dx$$.

Answer 3

I would write things down in a slightly different way: $$\int_0^{+\infty} e^{-x}\cos x \,dx = \int_0^{+\infty} e^{-x}\,d(\sin x) = \lim_{R \to +\infty} e^{-x}\sin x|_0^R + \int_0^{\infty}e^{-x}\sin x\,dx.$$ Now observe that $\lim_{R \to +\infty} e^{-R}\sin R = 0$ and $\sin 0 = 0$, so $\lim_{R \to +\infty} e^{-x}\sin x|_0^R = 0$ and you are done.

Answer 4

The difference is $\int_0^\infty e^{-x}(\cos x -\sin x)dx=[e^{-x}\sin x]_0^\infty =0$.