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Hint: To show that we have convergence for $\alpha\gt 0$, use the inequality $\log(1+x)\lt x$ if $x\gt 0$. To show we do not have convergence for $\alpha\le 0$, show first that we do not have convergence at $\alpha=0$. This can be done in various ways. One way is to use an estimate of $\log(1+1/k)$, Another is to note that $\log(1+1/k)=\log(k+1)-\log k$ ...

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HINT: We provide herein only an outline of the approach to evaluating the integral of interest using contour integration. Step 1: First note that $\frac{\sin z}{z}$ is an even function. Therefore, $$\int_0^\infty\frac{\sin^5 x}{x^5}\,dx=\frac12 \int_{-\infty}^\infty\frac{\sin^5 x}{x^5}\,dx$$ Step 2: Use Euler's Identity along with the Binomial ...

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One of the method is to use Integration By Parts... Dirichlet Integral: $$\displaystyle\int_0^\infty \dfrac{\sin x}x \,dx= \dfrac{\pi}2$$ $\displaystyle \int_0^\infty \dfrac{\sin^3x}xdx=\dfrac\pi4$ : We know $\sin^3x = \dfrac{3\sin x -\sin3x}4$ \begin{align*} \int_0^\infty \dfrac{\sin^3x}xdx &=\dfrac14 \int_0^\infty\dfrac{3\sin ...

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