Find values of $a$ for which the integral $$\int^{\infty}_{0}e^{-at}\sin(7t)dt$$ converges
What i try
$$\int^{\infty}_{0}e^{-at}\sin(7t)dt$$
$$=\frac{1}{a^2+49}\bigg(-e^{-at}a\sin(7t)-7e^{-at}\cos(7t\bigg)\bigg|^{\infty}_{0}=\frac{7}{a^2+49}$$
The integral is converges for all real $a$
What i have done above is right. If not then please tell me how do i solve it. Thanks
That converges for all $a>0$ since then we have that $$e^{-at}|\sin 7t|\le e^{-at}.$$
You can directly see that it does not when $a\le 0.$
For $z\in\Bbb C$,$$\int_0^\infty e^{-zt}dt=\frac1z(1-\lim_{t\to\infty}e^{-zt}),$$ which converges iff$$0=\lim_{t\to\infty}|e^{-zt}|=\lim_{t\to\infty}e^{-(\Re z)t},$$i.e. $\Re z>0$. So $\Im\int_0^\infty e^{-(a-7i)t}dt$ converges iff $\Re a>0$. If $a\in\Bbb R$, this simplifies to $a>0$.
