Evaluate this integral $\int ^\infty _0 \frac{z^{a-1}}{z^2+b}dz$

Question

Evaluate this integral $$\int ^\infty _0 \frac{z^{a-1}}{z^2+b}dz \;\;\;\; 0<a<2,b>0$$

My attempt:

here the zeros of $z^2+b=0 \Rightarrow z=\pm bi$

and $z=bi$ is the upper Half plan and it is only pole of order one so,

$Res_{z=bi}f(z)=\lim _{z\to bi} \{(z-ib)\frac{z^{a-1}}{z^2+b}\}\\ =\lim _{z\to bi} \{(z-ib)\frac{z^{a-1}}{z+ib}\}$

is i a m going right way..need help thank you..........

You'll need to consider various cases of $a$... – Matthew Cassell Mar 16 '18 at 02:34 — Matthew Cassell, Mar 16 '18 at 02:34
is that $\mathrm{d}x$ or $\mathrm{d}z$? – robjohn Mar 16 '18 at 02:37 — robjohn, Mar 16 '18 at 02:37
@robjohn.sorryyyyyy i edied – Inverse Problem Mar 16 '18 at 02:38 — Inverse Problem, Mar 16 '18 at 02:38

score 2 · Answer 1 · answered Mar 16 '18 at 02:42

2

Hint: $$ \int_0^\infty\frac{z^{a-1}}{z^2+b}\,\mathrm{d}z $$ Substitute $z=\sqrt{bx}$, then look at this answer.

answered Mar 16 '18 at 02:42

robjohn

345,667

score 2 · Accepted Answer · answered Mar 16 '18 at 18:45

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \left.\int_{0}^{\infty}{z^{a - 1} \over z^{2} + b}\,\dd z \,\right\vert_{\large\ {0\ <\ \Re\pars{a}\ <\ 2 \atop b\ >\ 0}} & \,\,\,\stackrel{z\ =\ b^{\large 1/2}\,t}{=}\,\,\, \int_{0}^{\infty}{b^{\pars{a - 1}/2}\,t^{a - 1} \over bt^{2} + b}\,b^{1/2} \,\dd t = b^{a/2 - 1}\int_{0}^{\infty}{t^{a - 1} \over t^{2} + 1}\,\dd t \\[5mm] & \stackrel{t^{2}\ \mapsto\ t}{=}\,\,\, {1 \over 2}\,b^{a/2 - 1}\int_{0}^{\infty}{t^{a/2 - 1} \over t + 1}\,\,\dd t \\[5mm] & = {1 \over 2}\,b^{a/2 - 1}\int_{0}^{\infty}t^{a/2 - 1}\int_{0}^{\infty}\expo{-\pars{t + 1}x}\,\dd x\,\dd t \\[5mm] & = {1 \over 2}\,b^{a/2 - 1}\int_{0}^{\infty}\expo{-x}\ \underbrace{\int_{0}^{\infty}t^{a/2 - 1}\expo{-xt}\,\dd t\,\dd x} _{\ds{\Gamma\pars{a/2} \over x^{a/2}}} \\[5mm] & = {1 \over 2}\,b^{a/2 - 1}\,\Gamma\pars{a \over 2} \int_{0}^{\infty}x^{-a/2}\expo{-x}\,\dd x \\[5mm] & = {1 \over 2}\,b^{a/2 - 1}\,\Gamma\pars{a \over 2}\Gamma\pars{-\,{a \over 2} + 1} = \bbx{{1 \over 2}\,b^{a/2 - 1}\,{\pi \over \sin\pars{\pi a/2}}} \end{align}

.......awesome sir tank you so much – Inverse Problem Mar 17 '18 at 01:59 — Inverse Problem, Mar 17 '18 at 01:59
@SureshPonnada You're welcome. Thanks. – Felix Marin Mar 17 '18 at 02:02 — Felix Marin, Mar 17 '18 at 02:02

Evaluate this integral $\int ^\infty _0 \frac{z^{a-1}}{z^2+b}dz$

2 Answers2