Evaluate this integral without using DogTeX

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

It is trivial to prove the equation above using DogTeX, but can you prove it without DogTeX?


This is a part of the set Formidable Series and Integrals

#Calculus

Note by Pi Han Goh
5 years, 2 months ago

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Comments

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

=120exsin(2x)xdx\displaystyle =\dfrac{1}{2} \int_{0}^{\infty} \dfrac{e^{-x} \sin(2x)}{\sqrt{x}} \mathrm{d}x

=14i0ex(e2ixe2ix)xdx\displaystyle = \dfrac{1}{4i} \int_{0}^{\infty} \dfrac{e^{-x} (e^{2ix} - e^{-2ix})}{\sqrt{x}} \mathrm{d}x

=AA4i\displaystyle = \dfrac{\text{A}-\text{A}^{*}}{4i}

=12(A)\displaystyle = \dfrac{1}{2} \Im (\text{A})

where A=0ex(12i)xdx\displaystyle \text{A} = \int_{0}^{\infty} \dfrac{e^{-x(1-2i)}}{\sqrt{x}} \mathrm{d}x

Note that,

Γ(t)=0xt1exdx\displaystyle \Gamma (t) = \int_{0}^{\infty} x^{t-1} e^{-x} \mathrm{d}x

Substituting xaxx \mapsto ax, we have,

Γ(t)=at0xt1eaxdx\displaystyle \Gamma(t) = a^{t} \int_{0}^{\infty} x^{t-1} e^{-ax} \mathrm{d}x

    A=Γ(12)112i\displaystyle \implies \text{A} = \Gamma \left( \dfrac{1}{2} \right) \dfrac{1}{\sqrt{1-2i}}

    I=π2(112i)\displaystyle \implies \text{I} = \dfrac{\sqrt{\pi}}{2} \Im \left( \dfrac{1}{\sqrt{1-2i}} \right)

=π254sin(12tan1(2)) \displaystyle = \boxed{\dfrac{\sqrt{\pi}}{2 \sqrt[4]{5}} \cdot \sin\left(\dfrac{1}{2} \tan^{-1}(2)\right)}

Ishan Singh - 5 years, 2 months ago

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Wonderful work as usual! +1

Pi Han Goh - 5 years, 2 months ago

@Ishan Singh good ishu... :)

Aman Rajput - 5 years, 2 months ago

Rajdeep Dhingra - 5 years, 2 months ago

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So cute!!!!

Nihar Mahajan - 5 years, 2 months ago

Marvellous solution! +1

Pi Han Goh - 5 years, 2 months ago

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LOL!

Swapnil Das - 5 years, 2 months ago
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