How to calculate this integration? [on hold]

Assume that βï¼‍αï¼‍0\betaï¼‍\alphaï¼‍0

How to calculate∫∞0e−αx2−e−βx2x2dx\int_{0}^{\infty}\frac{e^{-\alpha x^{2}}-e^{-\beta x^{2}}}{x^{2}}dxï¼ں

Many thanks for your help

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2 Answers
2

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First rewrite the integrand as an integral.
Idef=∫∞0e−ax2−e−bx2x2dx=∫∞0∫bae−ux2dudx\mathcal{I}\stackrel{def}{=}\int_0^\infty \frac{e^{-ax^2} – e^{-bx^2}}{x^2}dx
= \int_0^\infty \int_a^b e^{-ux^2} du dx
Since e−ux2e^{-ux^2} is non-negative, we can switch order of integration
and get
I=∫ba∫∞0e−ux2dxdu=∫ba√π2√udu=√πb−√πa\mathcal{I} = \int_a^b \int_0^\infty e^{-ux^2} dx du
= \int_a^b \frac{\sqrt{\pi}}{2\sqrt{u}} du = \sqrt{\pi b} – \sqrt{\pi a}

Suppose fix α\alpha consider β\beta as a parameter, i.e. look at the function
f(β)=∫∞0e−αx2−e−βx2x2 dx.\begin{align}
f(\beta) = \int^\infty_0 \frac{e^{-\alpha x^2}-e^{-\beta x^2}}{x^2}\ dx.
\end{align}
Observe
f′(β)=∫∞0x2e−βx2x2 dx=∫∞0e−βx2 dx=12√πβ\begin{align}
f'(\beta) = \int^\infty_0 \frac{x^2e^{-\beta x^2}}{x^2}\ dx = \int^\infty_0 e^{-\beta x^2}\ dx = \frac{1}{2}\sqrt{\frac{\pi}{\beta}}
\end{align}
which means
f(β)=C+√πβ\begin{align}
f(\beta) = C+\sqrt{\pi\beta}
\end{align}
Set β=α\beta= \alpha, we have that
f(α)=C+√πα=0  ⇒  C=−√πα.\begin{align}
f(\alpha)=C+\sqrt{\pi\alpha}=0 \ \ \Rightarrow \ \ C=-\sqrt{\pi\alpha}.
\end{align}
Thus, it follows
∫∞0e−αx2−e−βx2x2 dx=√πβ−√πα\begin{align}
\int^\infty_0 \frac{e^{-\alpha x^2}-e^{-\beta x^2}}{x^2}\ dx = \sqrt{\pi\beta}-\sqrt{\pi\alpha}
\end{align}