Where did I go wrong evaluating this integral?

Question

I wanted to see if complex analysis could help me evaluate the following integral: $$I=\int_{0}^{\infty} \cos(x) e^{−x} dx $$ It definitely converges due to the exponentially decaying function.

I began by substituting $\cos(x)$ with $\frac{e^{ix}+ e^{−ix}}{2}$: $$=\int_{0}^{\infty} \frac{e^{ix}+ e^{−ix}}{2} e^{−x} dx $$ Factor the $\frac{1}{2}$: $$=\frac{1}{2}\int_{0}^{\infty} (e^{ix}+ e^{−ix}) e^{−x} dx $$ Distribute $e^{-x}$: $$= \frac{1}{2} \int_{0}^{\infty}e^{ix}e^{-x} + e^{-ix}e^{-x} dx $$ Use the exponents rule: $$= \frac{1}{2} \int_{0}^{\infty}e^{ix-x} + e^{-ix-x} dx $$ Factor: $$= \frac{1}{2} \int_{0}^{\infty}e^{x(i-1)} + e^{-x(i+1)} dx$$ Using a simple u-substitution we evaluate the integral to be: $$= \frac{1}{2}\left [\frac{e^{ix-x}}{i-1} + \frac{e^{-ix-x}}{i+1}\right ]_{0}^{\infty}$$ Factor out $e^{-x}$: $$= \frac{1}{2}\left [\left (\frac{e^{ix}}{i-1} + \frac{e^{-ix}}{i+1} \right ) e^{-x}\right ]_{0}^{\infty}$$ Combine fractions: $$= \frac{1}{2}\left [\left (\frac{(i+1)e^{ix}+(i-1)e^{-ix}}{-2} \right ) e^{-x}\right ]_{0}^{\infty}$$ Expand: $$= \frac{1}{2}\left [\left (\frac{ie^{ix}+e^{ix}+ie^{-ix}-e^{-ix}}{-2} \right ) e^{-x}\right ]_{0}^{\infty}$$ Combine terms with an $i$ coefficient: $$= \frac{1}{2}\left [\left (\frac{(ie^{ix}+ie^{-ix})+(e^{ix}-e^{-ix})}{-2} \right ) e^{-x}\right ]_{0}^{\infty}$$ Factor out an $i$: $$= \frac{1}{2}\left [\left (\frac{i(e^{ix}+e^{-ix})+(e^{ix}-e^{-ix})}{-2} \right ) e^{-x}\right ]_{0}^{\infty}$$ Split the fraction: $$= \frac{1}{2}\left [\left (\frac{i(e^{ix}+e^{-ix})}{-2}+\frac{(e^{ix}-e^{-ix})}{-2} \right ) e^{-x}\right ]_{0}^{\infty}$$ Factor the negative: $$= -\frac{1}{2}\left [\left (i\frac{e^{ix}+e^{-ix}}{2}+\frac{e^{ix}-e^{-ix}}{2} \right ) e^{-x}\right ]_{0}^{\infty}$$ The exponential definition of the cosine function is: $$\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$$ Multiply by $i$: $$i\cos(x)=i\frac{e^{ix}+e^{-ix}}{2}$$

The exponential definition of the sine function is: $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$ Multiply by $i$: $$i\sin(x)=\frac{e^{ix}-e^{-ix}}{2}$$ We can substitute these back into our problem and get: $$= -\frac{1}{2}\left [\left ( i\cos(x)+i\sin(x) \right ) e^{-x}\right ]_{0}^{\infty}$$ Factor put the $i$: $$= -\frac{i}{2}\left [\left ( \cos(x)+\sin(x) \right ) e^{-x}\right ]_{0}^{\infty}$$ Let: $$A=\lim_{x \rightarrow \infty}(\cos(x)+\sin(x))$$ Since $-1 \le \sin(x) \le 1$ for $-\infty \le x \le \infty$ and $-1 \le \sin(x) \le 1$ for $-\infty \le x \le \infty$: $$-2 \leq \cos(x)+\sin(x) \leq 2$$ This implies: $$-2 \leq A \leq 2$$ As $x$ goes to infinity $e^{-x}$ goes to $0$ Evaluate bounds of integration: $$= -\frac{i}{2}\left [A \cdot e^{-\infty} -(\cos(0)+\sin(0))\cdot e^0\right ]$$ $$= -\frac{i}{2}\left [A \cdot 0 -(1 + 0)\cdot 1\right ]$$ $$= -\frac{i}{2}\left [0 - 1\right ]$$ $$= -\frac{i}{2}(- 1)$$ $$= -\frac{i}{2}$$ So: $$\int_{0}^{\infty} \cos(x) e^{−x} dx = \frac{i}{2}$$ It's obvious that I went wrong somewhere, I just cant figure out where!

Is there a reason you want to avoid integration by parts? This looks like an ideal candidate for IbP. — Clayton, Jul 18 '19 at 01:45
'Using a simple u-substitution we evaluate the integral to be' - there should be a minus sign there. Also, at this step you can just sub in the limits. For infinity both of them will be zero. — Jakobian, Jul 18 '19 at 01:48
@mathworker21 oooops, I actually got 1/2 when i fixed the sign error — Joshua Pasa, Jul 18 '19 at 02:03
At $I=\frac12\left[\frac{e^{x(i-1)}}{i-1}-\frac{e^{-x(i+1)}}{i+1}\right]_0^\infty$, why not just plug in your bounds? Doing so gives the correct answer and in much less steps... — Simply Beautiful Art, Jul 18 '19 at 02:07
@Clayton I couldnt use it because $\cos(x)$ can be differentiated infinatley many times and never reaches a constant $e^-x$ also does this, so IbP wont work. — Joshua Pasa, Jul 18 '19 at 02:09
@SimplyBeautifulArt wanted to remove the imaginary part so I expanded it so it could be written as a combination of sin and cos — Joshua Pasa, Jul 18 '19 at 02:11
You are kidding, right? Integration by parts works wonders here. — Clayton, Jul 18 '19 at 05:26
Yea it does work, didn't attempt it. I didn't think about solving for $\int e^x cos(x) dx$ and then dividing the expression by two. — Joshua Pasa, Jul 18 '19 at 13:02

user317176 · Answer 1 · 2019-07-18T02:26:54.447

You look fine up to here

$\frac{1}{2}\left [\left (\frac{i(e^{ix}-e^{-ix})}{-2}+\frac{(e^{ix}+e^{-ix})}{-2} \right ) e^{-x}\right ]_{0}^{\infty}$

At this point, I would say:

$\frac{1}{2}\left [\left (\frac{i(e^{ix}-e^{-ix})}{2i^2} - \frac{(e^{ix}+e^{-ix})}{2} \right ) e^{-x}\right ]_{0}^{\infty}\\ \frac{1}{2}\left [\left (\frac{(e^{ix}-e^{-ix})}{2i} - \frac{(e^{ix}+e^{-ix})}{2} \right ) e^{-x}\right ]_{0}^{\infty}$

Move back to trig functions. It looks like you added an extra $i$ factor at this step.

$\frac{(e^{ix}-e^{-ix})}{2i} = \sin x, \frac{(e^{ix}+e^{-ix})}{2} = \cos x\\ \frac{1}{2}[(\sin x - \cos x ) e^{-x}]_{0}^{\infty}\\ $

and evaluate.

$-\frac{1}{2}[(\sin 0 - \cos 0 ) = \frac {1}{2}$

Alternatively, at a much earlier step

$\frac{1}{2}\left [\frac{e^{(ix-x}}{-1+i} + \frac{e^{-ix-x}}{-1-i}\right ]_{0}^{\infty}\\ \frac{1}{2}\left [\frac{e^{-x}(\cos x + i\sin x)}{-1+i} + \frac{e^{-x}(\cos x -i\sin x)}{-1-i}\right ]_{0}^{\infty}\\ \frac{1}{2}e^{-x}\left [\frac{(-1-i)(\cos x + i\sin x) + (-1+i)(\cos x -i\sin x)}{(-1+i)(-1-i)}\right ]_{0}^{\infty}\\ \frac{1}{2}e^{-x}\left [\frac{-\cos x -i\sin x - i\cos x + \sin x -\cos x + i\sin x + i\cos x + \sin x}{2}\right ]_{0}^{\infty}\\ \frac{1}{2}e^{-x}\left [-\cos x + \sin x\right ]_{0}^{\infty}\\ $

score 2 · Accepted Answer · answered Jul 18 '19 at 02:40

2

Hint.

Why not integrating instead

$$ \mathcal{Re}\left(\int_0^{\infty}e^{ix}e^{-x} dx\right) $$

Easily we find

$$ \int_0^{\infty}e^{ix}e^{-x} dx = \left.\frac{1}{i-1}e^{(i-1)x}\right]_0^{\infty} = \frac{1}{1-i} = \frac{1}{2}+\frac i2 $$

answered Jul 18 '19 at 02:40

Cesareo

33,252

That's really elegant, I cant believe I didn't think of that – Joshua Pasa Jul 18 '19 at 02:45

Where did I go wrong evaluating this integral?

2 Answers2