Show that $\int_0^\infty e^{-x}\cos x \text{dx}=\int_0^\infty e^{-x} \sin x \text{dx}$ 
Show that $\int_0^\infty e^{-x}\cos x \ \text{d}x=\int_0^\infty e^{-x} \sin x \ \text{d}x$ using integration by parts.

For the LHS, I got:
$-e^{-x} \cos x-\int e^{-x} \sin x \ \text{d}x$
I'm not sure how to show that the RHS integral is equal to this...
 A: firstly you look the LHS:
$$\int \:e^{-x}\cos \left(x\right)dx$$
Now you substitute with $u=-x:\quad \quad du=-1dx,\:\quad \:dx=\left(-1\right)du$:
So,
\begin{align*}
=\int \:e^u\cos \left(x\right)\left(-1\right)du \\
=\int \:-e^u\cos \left(x\right)du \\
\end{align*}
Substitute: $u=-x\quad \Rightarrow \quad \:x=-u$
$$\Rightarrow =-\int \:e^u\cos \left(-u\right)du =-\int \:e^u\cos \left(u\right)du$$
$\mathrm{Apply\:Integration\:By\:Parts}:\quad \int \:uv'=uv-\int \:u'v$
$$=-\left(e^u\sin \left(u\right)-\int \:e^u\sin \left(u\right)du\right)$$
$$=-\left(e^u\sin \left(u\right)-\left(e^u\left(-\cos \left(u\right)\right)-\int \:e^u\left(-\cos \left(u\right)\right)du\right)\right)$$
$$=-\left(e^u\sin \left(u\right)-\left(-e^u\cos \left(u\right)-\int \:-e^u\cos \left(u\right)du\right)\right)$$
$\mathrm{Isolate}\:\int \cos \left(u\right)e^udu$
$$=-\frac{e^u\left(\sin \left(u\right)+\cos \left(u\right)\right)}{2}$$
$\mathrm{Substitute\:back}\:u=-x$ , simplify and add a constant:
$$=-\frac{e^{-x}\left(\cos \left(x\right)-\sin \left(x\right)\right)}{2}+C$$
Now calculate the limes for $x\to\infty$:
$$\lim _{x\to \infty \:}\left(-\frac{e^{-x}\left(\cos \left(x\right)-\sin \left(x\right)\right)}{2}\right)=0$$
Now, the limes for $x\to 0+$:
$$\lim _{x\to \:0+}\left(-\frac{e^{-x}\left(\cos \left(x\right)-\sin \left(x\right)\right)}{2}\right)=-\frac{1}{2}$$
$$\Rightarrow 0-\left(-\frac{1}{2}\right) = \frac{1}{2}$$
I hope you can do this for the RHS :)
A: Here is an answer that does not use integration by parts:
$$\int_0^\infty e^{-x} e^{ix} dx = \int_0^\infty e^{-x} \cos x dx + i \int_0^\infty e^{-x} \sin x dx = {1 \over i-1} e^{(i-1)x} \big|_0^\infty = {1 \over 2} (1+i)$$ hence
$$\int_0^\infty e^{-x} \cos x dx = \int_0^\infty e^{-x} \sin x dx = {1 \over 2}$$
A: Integrating by parts twice for each of the following gives
\begin{align}
\int_0^\infty e^{-x} \cos x \, dx &=\left.-\frac 12e^{-x}(\sin x+\cos x) \right\vert_0^\infty= \frac 12 \\
 \int_0^\infty e^{-x} \sin x \, dx &= \left.\frac 12e^{-x}(\sin x-\cos x) \right\vert_0^\infty = \frac 12
\end{align}
A: $$\int_{0}^{+\infty}(\cos x-\sin x) e^{-x}\,dx = \left.e^{-x}\sin x\right|_{0}^{+\infty} = 0.$$
A: You don't need to compute the integrals to show that they're equal. First of all, for any bounded function $f$ on $[0,\infty)$, the integral
$$
\int_0^\infty e^{-x}f(x)\,dx
$$
converges absolutely, because, if $|f(x)|\le k$, we have $|e^{-x}f(x)|\le ke^{-x}$ and
$$
\int_0^\infty e^{-x}\,dx
$$
converges.
Now, integrating by parts,
$$
\int_0^{\infty}e^{-x}\cos x\,dx=
[e^{-x}\sin x]_0^{\infty}-\int_0^\infty (-e^{-x})\sin x\,dx=
\int_0^\infty e^{-x}\sin x\,dx
$$
because, clearly,
$$
\lim_{x\to\infty}e^{-x}\sin x=0.
$$
