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Problem: Evaluate $$\int_{0}^{\infty} \dfrac{\sin^3{x}}{x \cdot e^x} dx=\dfrac{A\pi}{B}-\dfrac{\tan^{-1} (C)}{D},$$ My attempt through Differentiation under the Integral Sign: Using $\sin^3x=\frac{3\sin x-\sin 3x}{4}$, we can rewrite the integral as $$ \frac{3}{4}\int_0^\infty \frac{e^{-x}\sin x}{x}\,dx-\frac{1}{4}\int_0^\infty \frac{e^{-x}\sin 3x}{x}\,dx $$ Now, consider $$ I(b)=\int_0^\infty \frac{e^{-x}\sin bx}{x}\,dx\qquad,\qquad\mbox{where}\,\,b>0 $$ Differentiating w.r.t. (b), we have $$ \begin{align} I'(b)&=\int_0^\infty e^{-x}\cos bx\,dx\\ \end{align} $$ Could somebody please show me how to solve this Improper Integral? I tried solving it by using Integration by Parts twice, but for some bizarre reason, I got the value as $0$. $$$$ Please help me. Thanks very much in advance!

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  • $\begingroup$ What are $a,b,c,d$? $\endgroup$ – user223391 May 31 '15 at 16:36
  • $\begingroup$ Oh, that's just the closed form of the Integral. $\endgroup$ – Ishan May 31 '15 at 16:37
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    $\begingroup$ @avid19: it looks like a problem from Brilliant. One is usually supposed to find $a,b,c,d$. $\endgroup$ – Jack D'Aurizio May 31 '15 at 16:38
  • $\begingroup$ Yeah, it is. I failed to solve this one:( $\endgroup$ – Ishan May 31 '15 at 16:39
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Hint. You may just write, for $b \in \mathbb{R}$, $$ \int_0^{\infty}e^{-x}\cos bx\: dx=\Re\int_0^{\infty}e^{-(1+ib)x} dx=\Re\left. \frac{e^{-(1+ib)x}}{-(1+ib)}\right|_0^{\infty}=\frac{1}{1+b^2}. $$ Alternatively, integrating by parts twice: $$ \begin{align} I(b)=\int_0^{\infty}e^{-x}\cos bx\: dx&=\left.-e^{-x}\cos bx\right|_0^{\infty}-b\int_0^{\infty}e^{-x}\sin bx\: dx\\\\ &=1-b\times\left(\left.-e^{-x}\sin bx\right|_0^{\infty}+b\int_0^{\infty}e^{-x}\cos bx\: dx\right)\\\\ &=1-b^2I(b) \end{align} $$ giving $\displaystyle (1+b^2)I(b)=1$ and $$ I(b)=\frac{1}{1+b^2}. $$

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  • $\begingroup$ Sorry Sir, but I don't know how to use complex numbers in Integrals.If it's not too much trouble Sir, could you please suggest a place from where I could learn how to use them? $\endgroup$ – Ishan May 31 '15 at 16:43
  • $\begingroup$ Sir, could this Improper Integral be done using Integration by Parts, please? $\endgroup$ – Ishan May 31 '15 at 16:44
  • $\begingroup$ @BetterWorld By parts? yes. I will edit my answer. Thanks. $\endgroup$ – Olivier Oloa May 31 '15 at 16:45
  • $\begingroup$ @BetterWorld Done. Hoping it is OK for you now :) $\endgroup$ – Olivier Oloa May 31 '15 at 16:51
  • $\begingroup$ Sir, if it's not too much trouble, could you just explain one last thing, Sir? Sir, how do we evaluate $$-e^{-x}\cos bx|_0^{\infty}$$ and $$-e^{-x}\sin bx|_0^{\infty}$$ I had thought it was through this method. $\endgroup$ – Ishan May 31 '15 at 16:57
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We may use: $$ \int_{0}^{+\infty}\frac{f(x)}{x\,e^{x}}\,dx = \int_{1}^{+\infty}(\mathcal{L}\,f)(t)\,dt \tag{1}$$ so we just need to find the Laplace transform of $\sin^3(x)$.

Pretty easy task through the triplication formulas, since $\mathcal{L}(\sin x)=\frac{1}{1+t^2}$ implies: $$ \mathcal{L}(\sin^3 x) = \frac{3}{4}\left(\frac{1}{t^2+1}-\frac{1}{t^2+9}\right)=\frac{6}{(1+t^2)(9+t^2)}\tag{2}$$ then: $$ \int_{0}^{+\infty}\frac{\sin^3 x}{x\,e^{x}}\,dx = \frac{3\pi}{16}-\frac{3}{4}\int_{1}^{+\infty}\frac{dt}{t^2+9}=\color{red}{\frac{3\pi}{16}-\frac{\arctan 3}{4}}.\tag{3}$$

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  • $\begingroup$ Again Sir, I'm really sorry Sir, but due to my exceedingly limited knowledge, I don't know Laplace transforms. Actually, I learn Integration for fun (according to my course of Education here in India, I'm to start Integration officially only next November). Really sorry Sir for having troubled you. $\endgroup$ – Ishan May 31 '15 at 16:50
  • $\begingroup$ @BetterWorld: I hope you just need the definition. With a little digging into Wikipedia, it won't be difficult to prove $(1)$. The Laplace transform of a function $f(x)$ is defined as: $$(\mathcal{L} f)(s)=\int_{0}^{+\infty} f(x)\, e^{-sx}\,dx.$$ $\endgroup$ – Jack D'Aurizio May 31 '15 at 16:52
  • $\begingroup$ Alright Sir. Thanks a lot! $\endgroup$ – Ishan May 31 '15 at 17:38

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