The question is for what value of $a$, would $\int_0^{\infty} x^a \cos(x) dx$ converge, and since $\cos(x)$ is periodic, I guess, if $a > 0$, then $x^a$ monotonically increasing so the integral will not converge, and if $a = 0$, the integral of $\cos x$ will not converge either but how can I analyze when $a < 0$? I tried some examples using Wolfram|Alpha and I believe it will converge for $-1 < a < 0$ but I don't know how to prove it. Thank you for any help.
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1The question Does the integral of this function exists for any real value of $\alpha$? contains hint-like answers to your question. I've closed as duplicate, alternate answers can be added there if necessary. The general idea is that if you have a continuous function whose convergence you are trying to investigate, then you must look at the asymptotic behaviour at both its endpoints and use appropriate convergence/non-convergence theorems at each end. – Sarvesh Ravichandran Iyer Sep 02 '21 at 00:01
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Interestingly enough, contours may be used to evaluate the integral when it converges, see here. – Sarvesh Ravichandran Iyer Sep 02 '21 at 00:06
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For $x>0$ you have $x^a \cos(x) \le x^\alpha$, which may help in the left hand end of a $-1 < a < 0$ proof – Henry Sep 02 '21 at 00:57
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Have you considered applying the Alternating Sum Test for convergence series? Notice
$$\int_0^\infty x^\alpha\cos(x)dx=\int_0^{\pi/2}x^\alpha\cos(x)dx+\sum_{n=0}^\infty\int_{n\pi+\pi/2}^{(n+1)\pi+\pi/2}x^\alpha\cos(x)dx \ (*)$$ Assuming all integrals on the right converge, the sum is alternating, hence it will converge if and only if $$\lim_{n\to\infty}\int_{n\pi+\pi/2}^{(n+1)\pi+\pi/2}x^\alpha\cos(x)dx=0$$ which happens precisely when $x^\alpha\to0$ as $x\to\infty$. This in turn happens if and only if $\alpha<0$. So that's one condition.
Note, we did assume that all integrals on the right of $(*)$ converge. This will be satisfied iff $$\int_0^{\pi/2}x^\alpha\cos(x)dx$$ converges. One can show using the squeeze theorem that the above will converge iff $$\int_0^{\pi/2}x^\alpha dx$$ converges, which happens precisely when $\alpha>-1$. Combining the two conditions we see the the R.H.S. (and hence the L.H.S.) of $(*)$ will converge iff $-1<\alpha<0$.
Melody
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3You also need to consider $\int\limits_0^{\pi/2}x^\alpha\cos(x)dx$, which puts a further constraint on $\alpha$ – Henry Sep 02 '21 at 00:54
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