finding convergence of integral having exponential and cosine terms

Question

Finding whether the series $$\int^{\infty}_{1}e^{x}\cos (x)\cdot x^{-\frac{1}{2}}dx$$ converges or diverges.

What i try::

$$I=\int^{\infty}_{1}e^{x}\cos(x)\cdot x^{-\frac{1}{2}}dx\leq \int^{\infty}_{1}e^{x}\cdot x^{-\frac{1}{2}}dx$$

Now put $x=t^2$ and $dx=2tdt$

$$I<2\int^{\infty}_{1}e^{t^2}dt<\infty$$

From series expansion.

But this does not show anything.

Hiw do i solve it. Help me please. Thanks.

I don't think the integral converges, because $e^x \cos x$ becomes both arbitrarily $<0$ and $>0$ and $e^x = \omega(\sqrt{x})$, — Alex, Jun 09 '20 at 22:51
Thanks alex but still i did not understand how can i prove that integral diverges. — jacky, Jun 09 '20 at 22:53

score 1 · Accepted Answer · answered Jun 09 '20 at 22:56

1

Due to the fact that $e^x \cos x$ diverges away from $0$ arbitrarily often, and $\frac{e^x}{\sqrt{x}}$ is an increasing function (since $e^x = \omega(\sqrt{x})$, the integrand doesn't $\to 0$ as $x \to \infty$. Hence, integral diverges.

answered Jun 09 '20 at 22:56

Alex

19,262

Thanks Alex for posting solution – jacky Jun 10 '20 at 00:23
1

@jacky Feel free to accept answers that solve your problems – Alex Jun 10 '20 at 01:04

finding convergence of integral having exponential and cosine terms

1 Answers1