Integrating $\int_{-\infty}^0e^x\sin(x)dx$

I ask if anyone could solve the following:

$$\int_{-\infty}^0e^x\sin(x)dx=?$$

I can visually see that it will converge and that it should be less than $1$:

$$\int_{-\infty}^0e^x\sin(x)dx<\int_{-\infty}^0e^xdx=1$$

But I am unsure what its exact value is.

Trying to find the definite integral by integrating by parts 4 times only results in $e^x\sin(x)$, which got me nowhere.

How should I evaluate this?

• +1 to this question. You seem genuinely interested in learning :) – user223391 Jan 13 '16 at 23:01

Integrating by parts two times you get \begin{align} \int e^x \cdot \sin x \quad dx & = e^x \cdot \sin x - \int e^x \cdot \cos x \quad dx\\ & = e^x \cdot \sin x - \left( \int e^x \cdot \cos x \quad dx \right)\\ & = e^x \cdot \sin x - \left( e^x\cdot \cos x + \int e^x \cdot \sin x\quad dx\right)\\ & = e^x \cdot \sin x - e^x\cdot \cos x - \int e^x \cdot \sin x\quad dx\\ & = e^x \cdot \left( \sin x - \cos x\right)- \int e^x \cdot \sin x\quad dx\\ \end{align} hence $$\int e^x \cdot \sin x \quad dx = e^x \cdot \left( \sin x - \cos x\right)- \int e^x \cdot \sin x\quad dx\\ \Rightarrow \int e^x \cdot \sin x \quad dx = \frac{e^x \cdot \left( \sin x - \cos x\right)}{2}$$

Finally finding $\int_{-\infty}^0e^x\sin(x)dx$ it's just a matter of substitutions \begin{align} \int_{-\infty}^0e^x\sin(x)dx &= \frac{e^x \cdot \left( \sin x - \cos x\right)}{2} \Big|_{-\infty} ^0\\ &= \frac{1 \cdot \left( 0 - 1\right)}{2} - \lim_{x \to -\infty} \frac{e^x \cdot \left( \sin x - \cos x\right)}{2}\\ &=-\frac{1}{2} - 0\\ &=-\frac{1}{2} \end{align}

• How did you get there? – Simply Beautiful Art Jan 13 '16 at 22:39
• Let $J = \int e^x \sin x \ d x$. Integrate by parts and we find $$J = e^x(\sin x - \cos x) - J$$ Then solve for $J$. At first it appears the procedure hasn't got us anywhere when in fact we have. – Simon S Jan 13 '16 at 22:45
• Oh, that's something I can really understand – Simply Beautiful Art Jan 13 '16 at 22:45

$$\int_{-\infty}^0e^x\sin(x)dx = \mbox{Im}\int_{-\infty}^0 e^{x(1+i)}dx =\mbox{Im}\left.\frac{1}{1+i}e^{x(1+i)}\right|_{-\infty}^0 =\mbox{Im}\frac{1}{1+i}=\mbox{Im}\frac{1-i}{2}=-\frac{1}{2}.$$

• I am not familiar with that method. Could you at least tell me what you did and what this method is called? – Simply Beautiful Art Jan 13 '16 at 22:40
• Treat the original integral as the imaginary part of another integral, which is much easier to integrate. – yhhuang Jan 13 '16 at 22:41
• Could you explain how you managed to find this other integral? – Simply Beautiful Art Jan 13 '16 at 22:42
• The other integral is just a simple one involving exponential. I added one more step. – yhhuang Jan 13 '16 at 22:44
• Well, I'm going to mark the other one as answered because it was easier for me. Thank you for your answer though, I will keep that in mind. :D – Simply Beautiful Art Jan 13 '16 at 22:46

It can easily be done by applying by parts twice: Let $$I=\int e^x \cdot \sin x \quad dx = e^x\sin x-\int e^x\cos x\,dx$$ for the second one apply by parts, $$\int e^x\cos x\,dx=e^x\cos x+I$$ so, $$I = \frac{1}{2} e^x (\sin x -\cos x)$$

In combination with yhhuang's answer, I think I have a similar explanation:

$$\int e^x\cdot\sin(x)dx=\int e^x\cdot\text{Im}(\cos(x)+i\sin(x))dx$$

$$=\int e^x\cdot\text{Im}(e^{xi})dx$$

$$=\text{Im}\int e^{x(1+i)}dx$$

I found this easier to understand.