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Problem. Calculate $$ \int_{0}^{\infty} e^{-x}\cos(x) \, dx $$

I was recommended to calculate it using limits, but firstly I have solved the integral and I got

$$ \int_{0}^{\infty} e^{-x}\cos(x) \, dx = \frac{\sin(x) - \cos(x)}{2e^x}. $$

I don't know how to calculate limit as $x$ tends to $\infty$.

Sangchul Lee
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Sofia Ibragimova
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    As $x$ tends to $\infty$, $ e^x \rightarrow \infty$, and as $-\sqrt{2}\le sin(x)-cos(x)\le \sqrt{2}$, your limit tends to 0. – Jake Stone Mar 08 '23 at 22:33
  • Got it! Thank you – Sofia Ibragimova Mar 08 '23 at 22:38
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    I think you mean you got $$ \int_0^x e^{-t}\cos(t),{\rm d}t=\frac{\sin x-\cos x}{2e^x}. $$ The dummy variable of definite integral does not appear outside of it, and a single-variable definite integral is a constant, not a function. – anon Mar 08 '23 at 23:56
  • @JudeCarter Note that $$ \int_0^{ + \infty } {{\rm e}^{ - x} \cos x,{\rm d}x} = \frac{1}{2}. $$ More generally, $$ \int_0^{ + \infty } {{\rm e}^{ - ax} \cos (bx),{\rm d}x} = \frac{a}{{a^2 + b^2 }} $$ for $\operatorname{Re}(a)>0$ and $b\in \mathbb{R}$. – Gary Mar 09 '23 at 01:05

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