How to integrate ${\int\limits_0^{\infty}\frac{x^{4}e^{-ax^2}}{x^2+k^2}\,\mathrm dx}$

Question

Integrate ${\displaystyle \int_0^{\infty}\frac{x^{4}e^{-ax^2}}{x^2+k^2}\,\mathrm dx}$.

Answer 1

As disscused in the comment section ,I am elaborting it, $$ I(a)=\int_{0}^{\infty} \frac{exp(-a^2x^2)}{(x^2+k^2)} dx = \frac{2 \pi}{k}\exp(-a^2k^2) erfc(ak) $$ now get second derivative of $I(a)$.Hence your answer will be, $$ I(a)\prime \prime =\int_{0}^{\infty} \frac{(k^4) exp(-a^2k^2) dx }{(x^2+k^2)} $$

Answer 2

First, we note that with the substitution $x=y\, k$ we get $$ I = \int_0^\infty \frac{x^4}{x^2+k^2}\operatorname{e} ^{-ax^2} \operatorname d x = k^2 \int_0^\infty \frac{y^4}{y^2+1}\operatorname{e} ^{-(ak^2)y^2} \operatorname d y $$

And now $$ \frac{y^4}{y^2+1}=\frac{y^4-1+1}{y^2+1}=\frac{(y^2+1)(y^2-1)+1}{y^2+1} = y^2 - 1 + \frac{1}{y^2+1} $$

And, if we put $b=ak^2$, the integral becomes $$ I = k^2 \left[ \int_0^\infty y^2\operatorname{e} ^{-by^2} \operatorname d y -\int_0^\infty \operatorname{e} ^{-by^2} \operatorname d y + \int_0^\infty \frac{\operatorname{e} ^{-by^2}}{y^2+1} \operatorname d y\right] = k^2 \left[ I_1-I_2+I_3\right] $$ with $$ \begin{split} I_1&=\int_0^\infty y^2\operatorname{e} ^{-by^2} \operatorname d y \\ I_2&= \int_0^\infty \operatorname{e} ^{-by^2} \operatorname d y \\ I_3&=\int_0^\infty \frac{\operatorname{e} ^{-by^2} }{y^2+1} \operatorname d y \end{split} $$

Hint for $I_2$: We know that $$ \int_0^\infty \operatorname e^{-x^2} \operatorname d x = \frac{\sqrt \pi}{2} $$ Can you continue from here?

[Result: $I_2 = \frac{1}{2}\sqrt{\frac{\pi}{b}}$]

Hint for $I_1$: We can write $$ y^2\operatorname{e} ^{-by^2} = -\frac{y}{2b} \cdot (-2yb \operatorname{e} ^{-by^2} ) = - \frac{y}{2b}\left(\operatorname{e} ^{-by^2} \right)' $$

Hence from integration by parts we get $\dots$

[Result: $I_1 = \frac{1}{4}\sqrt{\frac{\pi}{b^3}}$]

Hint for $I_3$:

Let $f(y)= \frac{1}{y^2+1}$ and $g(y)=\operatorname{e} ^{-by^2}$ and let $\mathcal F$ be the Fourier transform operator, then we have : $$ I_3 = \int_0^\infty \frac{\operatorname{e} ^{-by^2} }{y^2+1} \operatorname d y = \frac{1}{2}\int_{-\infty}^{+\infty} \frac{\operatorname{e} ^{-by^2} }{y^2+1} \operatorname{e}^{-\operatorname i 0 y} \operatorname d y = \frac{1}{2}\mathcal F (fg) (0) $$ Let $$ \mathcal F (fg) (k) = \int_{-\infty}^{+\infty} \frac{\operatorname{e} ^{-by^2} }{y^2+1} \operatorname{e}^{-\operatorname i k y} \operatorname d y $$ We remember that Fourier transform of product is the convolution of Fourier trasforms, in formulas: $$ \mathcal{F}(fg)(k) = \frac{1}{\sqrt{2\pi}} \mathcal{F}(f) \star \mathcal{F}(g) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} \mathcal{F}f(y) \, \mathcal{F}g (k-y) \operatorname d y $$ Now we remmeber that $$ \mathcal F f = \mathcal F \operatorname{e}^{-by^2} = \frac{1}{\sqrt {2b}} \operatorname{e}^{-\frac{k^2}{4b}} $$

and that $$ \mathcal F g = \mathcal F \, \frac{1}{1+y^2} =\sqrt{ \frac{2}{\pi}} \operatorname e ^{-|x|} $$ So the integral $I_3$ becames $$ \begin{split} I_3 &= \frac{1}{2} \mathcal{F}(fg)(0) = \frac{1}{2}\int_{-\infty}^{+\infty} \mathcal{F}f(y) \, \mathcal{F}g (-y) \operatorname d y\\ &= \frac{1}{2}\int_{-\infty}^{+\infty} \frac{1}{\sqrt {2b}} \operatorname{e}^{-\frac{y^2}{4b}} \sqrt{ \frac{2}{\pi}} \operatorname e ^{-|y|} \operatorname d y\\ &= \frac{1}{\sqrt{b \pi}} \int_0^{+\infty} \operatorname{e}^{-\frac{y^2}{4b}-y} \operatorname d y \end{split} $$

Can you continue from here? (Hint: complete the square of the exponent)