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I have to calculate $\displaystyle\int_0^{\infty} \dfrac{\sqrt{x}}{x^2+a^2} (a>0)$ using complex integral.

Let $f(z)=\dfrac{z^{\frac{1}{2}}}{z^2+a^2}$.

Integral Contour ($r<R$);

$C_1 : [r,R], z=t,t:r\to R$

$C_2$ : Circle that has radius $R, z=Re^{i\theta}, \theta : 0\to 2\pi$

$C_3$ : $[R,r], z=te^{2\pi i}, t: R\to r$

$C_4$ : Circle that has raius $r, z= re^{i\theta}, \theta : 2\pi \to 0$

\begin{equation} \displaystyle\int_{C_2} f(z) dz, \displaystyle\int_{C_4} f(z) dz \to 0\,\,({r\to 0, R\to \infty} ) \end{equation}

\begin{align} &\displaystyle\int_{C_1} f(z) dz \to \displaystyle\int_0^{\infty} \dfrac{\sqrt{x}}{x^2+a^2} dx.\\ &\displaystyle\int_{C_3} f(z) dz \to -e^{i\pi}\displaystyle\int_0^{\infty} \dfrac{\sqrt{x}}{x^2+a^2} dx =\displaystyle\int_0^{\infty} \dfrac{\sqrt{x}}{x^2+a^2} dx. \end{align}

From Residue Theorem, \begin{align} 2\displaystyle\int_0^{\infty} \dfrac{\sqrt{x}}{x^2+a^2} dx =2\pi i[\text{Res}(f, ai)+\text{Res}(f, -ai)] \end{align}

\begin{align} &\text{Res}(f, ai) =\dfrac{(ai)^{\frac{1}{2}}}{2a i} =\dfrac{e^{\frac{\pi i}{4}}}{2\sqrt{a}i}\\ &\text{Res}(f, -ai) =\dfrac{(-ai)^{\frac{1}{2}}}{-2a i} =-\dfrac{e^{-\frac{\pi i}{4}}}{2\sqrt{a}i}\\ \end{align}

Therefore, \begin{align} \displaystyle\int_0^{\infty} \dfrac{\sqrt{x}}{x^2+a^2} dx =\pi i \left[\dfrac{e^{\frac{\pi i}{4}}}{2\sqrt{a}i}-\dfrac{e^{-\frac{\pi i}{4}}}{2\sqrt{a}i}\right] =\dfrac{\pi }{\sqrt{2a}}i. \end{align} So $\displaystyle\int_0^{\infty} \dfrac{\sqrt{x}}{x^2+a^2} dx=0.$

But answer is $\displaystyle\int_0^{\infty} \dfrac{\sqrt{x}}{x^2+a^2} dx=\dfrac{\pi}{\sqrt{2a}}$. Maybe, my calculation is wrong, but I cannot find where I mistook.

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1 Answer 1

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Because you have cut along the positive real axis, the second residue is $-e^{3 \pi i /4}/(2 \sqrt ai)$ leading to, \begin{align} 2 \int_0^\infty \frac{\sqrt{x}}{x^2 + a^2} dx &= \frac{2\pi i}{2\sqrt a i} \Big( e^{\pi i/4} - e^{3 \pi i / 4} \Big) \\ &=\frac{\pi}{\sqrt{a}} \cdot \frac{2}{\sqrt 2} \end{align}

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