\[\large \int_{0}^{\infty} \frac{e^{-x} \sin(x)\cos(x)}{\sqrt{x}}\,dx = \frac{\sqrt{\pi}}{2\cdot \sqrt[4]{5}} \cdot \sin\left(\frac{1}{2} \tan^{-1}(2)\right)\]

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This is a part of the set Formidable Series and Integrals

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## Comments

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TopNewestLet \(\displaystyle \text{I} = \int_{0}^{\infty} \dfrac{e^{-x} \sin(x) \cos(x)}{\sqrt{x}} \mathrm{d}x\)

\(\displaystyle =\dfrac{1}{2} \int_{0}^{\infty} \dfrac{e^{-x} \sin(2x)}{\sqrt{x}} \mathrm{d}x \)

\(\displaystyle = \dfrac{1}{4i} \int_{0}^{\infty} \dfrac{e^{-x} (e^{2ix} - e^{-2ix})}{\sqrt{x}} \mathrm{d}x \)

\(\displaystyle = \dfrac{\text{A}-\text{A}^{*}}{4i}\)

\(\displaystyle = \dfrac{1}{2} \Im (\text{A})\)

where \(\displaystyle \text{A} = \int_{0}^{\infty} \dfrac{e^{-x(1-2i)}}{\sqrt{x}} \mathrm{d}x \)

Note that,

\(\displaystyle \Gamma (t) = \int_{0}^{\infty} x^{t-1} e^{-x} \mathrm{d}x \)

Substituting \(x \mapsto ax\), we have,

\(\displaystyle \Gamma(t) = a^{t} \int_{0}^{\infty} x^{t-1} e^{-ax} \mathrm{d}x\)

\(\displaystyle \implies \text{A} = \Gamma \left( \dfrac{1}{2} \right) \dfrac{1}{\sqrt{1-2i}} \)

\(\displaystyle \implies \text{I} = \dfrac{\sqrt{\pi}}{2} \Im \left( \dfrac{1}{\sqrt{1-2i}} \right) \)

\( \displaystyle = \boxed{\dfrac{\sqrt{\pi}}{2 \sqrt[4]{5}} \cdot \sin\left(\dfrac{1}{2} \tan^{-1}(2)\right)} \)

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Wonderful work as usual! +1

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@Ishan Singh good ishu... :)

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So cute!!!!

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Marvellous solution! +1

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LOL!

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