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I hope I can find closed form solution for two following definite integrals. Unfortunately I don't have Mathematica and I can't find similar integrals in Tables. Can any one help me please?

\begin{equation} \int_0^\infty \frac{e^{-x}}{x^{1/2}+a x^{3/2}} dx \end{equation}

\begin{equation} \int_0^\infty \frac{xe^{-x}}{x^{1/2}(1+bx)^2} dx \end{equation}

Alireza
  • 553

2 Answers2

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Here is a solution for the first integral \begin{align} \tag{a} \int\limits_{0}^{\infty} \frac{\mathrm{e}^{-x}}{x^{1/2}+ax^{3/2}} \mathrm{d}x &= 2\int\limits_{0}^{\infty} \frac{z\mathrm{e}^{-z^{2}}}{z+az^{3}} \mathrm{d}z \\ \tag{b} &= \frac{2}{\sqrt{a}} \int\limits_{0}^{\infty} \mathrm{e}^{-y^{2}/a} \frac{1}{1+y^{2}} \mathrm{d}y \\ \tag{c} &= \frac{\pi}{\sqrt{a}} \mathrm{e}^{1/a} \mathrm{erfc}\left(\frac{1}{\sqrt{a}}\right) \end{align}

Notes:

a. $x=z^{2}$

b. $y^{2}=az^{2}$

c. From DLMF, we have the following integral definition of the complementary error function \begin{equation} \mathrm{erfc}(z) = \frac{2}{\pi} \mathrm{e}^{-z^{2}} \int\limits_{0}^{\infty} \mathrm{e}^{-z^{2}t^{2}} \frac{1}{t^{2} + 1} \mathrm{d}t \end{equation}

Addendum

For a proof of the above equation, see Show $\frac{2}{\pi} \mathrm{exp}(-z^{2}) \int_{0}^{\infty} \mathrm{exp}(-z^{2}x^{2}) \frac{1}{x^{2}+1} \mathrm{d}x = \mathrm{erfc}(z)$

Community
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poweierstrass
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The second integral is calculated in the same fashion: \begin{eqnarray} &&\int\limits_0^\infty \frac{x e^{-x}}{\sqrt{x} (1+b x)^2} dx \underbrace{=}_{z=\sqrt{x}}\\ &&\int\limits_0^\infty \frac{z^2}{(1+b z^2)^2} 2 e^{-z^2} dz=\\ &&\int\limits_0^\infty \left(\frac{1}{b} \frac{1}{1+b z^2} - \frac{1}{b} \frac{1}{1+b z^2)^2} \right) 2 e^{-z^2} dz=\\ &&\frac{2}{b^{3/2}} \left( \int\limits_0^\infty \frac{e^{-\frac{z^2}{b}}}{1+z^2}dz - \int\limits_0^\infty \frac{e^{-\frac{z^2}{b}}}{(1+z^2)^2}dz\right)=\\ &&\frac{2}{b^{3/2}} \left( \frac{\pi}{2} \exp(\frac{1}{b}) erfc(\frac{1}{\sqrt{b}}) - \int\limits_0^\infty \frac{e^{-\frac{z^2}{b}}}{(1+z^2)^2}dz\right)=\\ &&\frac{2}{b^{3/2}} \left( \frac{\pi}{2} \exp(\frac{1}{b}) erfc(\frac{1}{\sqrt{b}}) - \int\limits_0^\infty \frac{1}{b} \left( \frac{z}{1+z^2} + \arctan(z)\right) z e^{-\frac{z^2}{b}} dz\right)=\\ && \frac{\pi}{b^{3/2}}\exp(\frac{1}{b})erfc(\frac{1}{\sqrt{b}}) - \frac{2}{b^{5/2}} \int\limits_0^\infty \left( 1-\frac{1}{1+z^2} + z \arctan(z) \right) e^{-\frac{z^2}{b}} dz=\\ &&\frac{\pi}{b^{3/2}}\exp(\frac{1}{b})erfc(\frac{1}{\sqrt{b}}) - \frac{2}{b^{5/2}}\left( \sqrt{b} \frac{\sqrt{\pi}}{2} - \frac{\pi}{2} \exp(\frac{1}{b}) erfc(\frac{1}{\sqrt{b}}) + \int\limits_0^\infty z e^{-\frac{z^2}{b}} \arctan(z) dz \right)=\\ &&\frac{\pi}{b^{3/2}}\exp(\frac{1}{b})erfc(\frac{1}{\sqrt{b}}) - \frac{2}{b^{5/2}}\left( \sqrt{b} \frac{\sqrt{\pi}}{2} - \frac{\pi}{2} \exp(\frac{1}{b}) erfc(\frac{1}{\sqrt{b}}) + \frac{b}{2} \int\limits_0^\infty e^{-\frac{z^2}{b}} \frac{1}{1+z^2} dz \right)=\\ &&\frac{\pi}{b^{3/2}}\exp(\frac{1}{b})erfc(\frac{1}{\sqrt{b}}) - \frac{2}{b^{5/2}}\left( \sqrt{b} \frac{\sqrt{\pi}}{2} - \frac{\pi}{2} \exp(\frac{1}{b}) erfc(\frac{1}{\sqrt{b}}) + \frac{b}{2} \frac{\pi}{2} \exp(\frac{1}{b}) erfc(\frac{1}{\sqrt{b}}) \right)=\\ && -\frac{\sqrt{\pi}}{b^2} + \frac{\pi}{2 b^{3/2}} \exp(\frac{1}{b}) erfc(\frac{1}{\sqrt{b}}) \left( 1+ \frac{2}{b} \right) \end{eqnarray}

Przemo
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