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While trying to solve a certain Laplace transform, this spicy integral developed.

$$ \int_0^\infty \frac{\sqrt x e^{-x}}{b^2 +x^2} dx $$

I am not sure how to approach this problem, and would appreciate any help. One attempt was to expand the exponential into a power series. This would lead to something like:

$$\sum_{n=0}^\infty \frac{{(-1)}^n}{n!} \int_0^\infty \frac{x^{n+1/2}}{b^2+x^2}dx$$

This integral looks like it could be approached with some complex analysis for $n < 3/2$ but it doesn't really make sense for all except for the first two terms. Am I missing something here?

I would appreciate any help solving this integral! Thanks.

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    $\begingroup$ Have you tried a contour integral of some sort? $\endgroup$ – Rivers McForge Jul 10 '20 at 21:00
  • $\begingroup$ Not successfully. I haven't managed to find a contour that isolates this term. Any suggestions? $\endgroup$ – guavas222 Jul 10 '20 at 21:10
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    $\begingroup$ In your attempt, interchanging the order of integration and summation is not always possible. Also, Mathematica tells that, for $b>0$, $$\int_{0}^{\infty}\frac{\sqrt{x}e^{-x}}{b^2+x^2}\,\mathrm{d}x=\frac{\pi}{\sqrt{2b}}\left[\cos(b)\left(1-2C\left(\sqrt{\frac{2b}{\pi}}\right)\right)+\sin(b)\left(1-2S\left(\sqrt{\frac{2b}{\pi}}\right)\right)\right],$$where $C(\cdot)$ and $S(\cdot)$ are the Fresnel integrals. $\endgroup$ – Sangchul Lee Jul 10 '20 at 21:43
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    $\begingroup$ My previous comment might be confusing because the parameter choice of the Fresnel integrals in Mathematica is different from that in Wikipedia article. A more transparent formula based on the above result is: $$\int_{0}^{\infty}\frac{\sqrt{x}e^{-x}}{b^2+x^2}\,\mathrm{d}x=\sqrt{\frac{\pi}{b}}\int_{0}^{\infty}\frac{\cos u}{\sqrt{u+b}}\,\mathrm{d}u.$$ $\endgroup$ – Sangchul Lee Jul 10 '20 at 21:55
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    $\begingroup$ Let $f(x) = \sqrt x$ and $g(x) = e^{-x}/(x^2 + b^2)$, then $$\int_0^\infty f(x) g(x) dx = \int_0^\infty \mathcal L[f](x) \mathcal L^{-1}[g](x) dx = \frac {\sqrt \pi} {2 b} \int_1^\infty \frac {\sin(b (x - 1))} {x^{3/2}} dx.$$ Substituting $x = u^2$ gives Fresnel integrals. $\endgroup$ – Maxim Jul 27 '20 at 19:56
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According to Maple it is (for $b > 0$) $$ 2\,\sqrt {\pi} \left( {\it LommelS2} \left( 1,1/2,b \right) -1 \right) $$

where the Lommel S2 function is defined here. So I don't think you're going to get an elementary answer.

EDIT: Writing $$ \frac{1}{b^2 + x^2} = \frac{i}{2 b (x + i b)} - \frac{i}{2 b (x - i b)}$$ I get something slightly more elementary: $$ -{\frac {\pi\,\sqrt {2} \left( \left( 1+i \right) {{\rm e}^{-ib} }{\rm erf} \left( \left( 1/2-i/2 \right) \sqrt {2}\sqrt {b}\right)+ \left( 1-i \right) {{\rm e}^{ib}}{\rm erf} \left( \left( 1/2+i/2 \right) \sqrt {2}\sqrt {b}\right)- \left( 1+i \right) {{\rm e}^{-ib}} - \left( 1-i \right) {{\rm e}^{ib}} \right) }{4 \sqrt {b}}} $$

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  • $\begingroup$ I can't say I am familiar with the LommelS2 function. I was expecting to see it reduce into some sort of gamma function. Is that reasonable? $\endgroup$ – guavas222 Jul 10 '20 at 21:12
  • $\begingroup$ Apparently not. $\endgroup$ – Robert Israel Jul 10 '20 at 21:15
  • $\begingroup$ This could definitely go somewhere useful (I am going to play with for a little bit). Thanks! $\endgroup$ – guavas222 Jul 10 '20 at 21:42
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Not sure how helpful this is but, we can turn the problem into a second order linear differential equation:

\begin{equation} I(a)=\int\limits_{0}^{+\infty} \frac{x^{k}e^{-ax}}{x^{2}+b^{2}}\,dx \end{equation}

for some positive real $k$. Using Leibniz's rule, the first and second derivatives with respect to $a$ are:

\begin{equation} I'(a)=\int\limits_{0}^{+\infty} \frac{x^{k}e^{-ax}(-x)}{x^{2}+b^{2}}\,dx \end{equation}

\begin{equation} I''(a)=\int\limits_{0}^{+\infty} \frac{x^{k}e^{-ax}x^{2}}{x^{2}+b^{2}}\,dx \end{equation}

In the second derivative, add and subtract $b^{2}$ in the $x^{2}$ term:

\begin{equation} I''(a)=\int\limits_{0}^{+\infty} \frac{x^{k}e^{-ax}(x^{2}+b^{2}-b^{2})}{x^{2}+b^{2}}\,dx \end{equation}

\begin{equation} I''(a)=\int\limits_{0}^{+\infty} x^{k}e^{-ax}\,dx-b^{2}\underbrace{\int\limits_{0}^{+\infty} \frac{x^{k}e^{-ax}}{x^{2}+b^{2}}\,dx}_{I(a)} \end{equation}

\begin{equation} I''(a)+b^{2}I(a)=\int\limits_{0}^{+\infty} x^{k}e^{-ax}\,dx \end{equation}

With the substitution $u=ax$, one can expressed the last remaining integral in terms of the gamma function:

\begin{equation} \int\limits_{0}^{+\infty} x^{k}e^{-ax}\,dx = \frac{\Gamma(k+1)}{a^{k+1}} = \frac{k!}{a^{k+1}} \end{equation}

Then, in order to compute $I(a)$, we need to solve the following ODE:

\begin{equation} I''(a)+b^{2}I(a)-\frac{k!}{a^{k+1}}=0 \end{equation}

For the integral in the question above, we have the case where $k=1/2$, then we would need to solve the following:

\begin{equation} I''(a)+b^{2}I(a)-\frac{a^{-\frac{3}{2}}\sqrt{\pi}}{2}=0 \end{equation}

The solution to this ODE given by WolframAlpha is quite nasty: https://www.wolframalpha.com/input/?i=y%27%27%28x%29%2Bcy%28x%29-%28%5Csqrt%28%5Cpi%29%2F2x%5E%28-3%2F2%29%29%3D0.

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    $\begingroup$ You're on the right track to solve this analytically. You would use variation of parameters on the inhomogeneous term. The trick is to be able to rewrite the integral in terms of known special functions like the exponential integral or cosine integral functions. $\endgroup$ – Ninad Munshi Jul 11 '20 at 16:33

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