# Evaluate this integral without using DogTeX

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

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 Note by Pi Han Goh
5 years, 3 months ago

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

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

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

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

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

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,

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

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

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

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

- 5 years, 3 months ago

Wonderful work as usual! +1

- 5 years, 3 months ago

@Ishan Singh good ishu... :)

- 5 years, 3 months ago - 5 years, 3 months ago

So cute!!!!

- 5 years, 3 months ago

Marvellous solution! +1

- 5 years, 3 months ago

LOL!

- 5 years, 3 months ago