How to calculate $$\int_0^{+\infty}\frac{e^{-ax}\sin^2(bx)}{x^2}dx$$ I tried to find $dI(a, b) = \frac{\partial I}{\partial a} + \frac{\partial I}{\partial b}$, but these partial derivative are not easy to calculate. Could you advise me how to start?
Calculation of $\int_0^{+\infty}\frac{e^{-ax}\sin^2(bx)}{x^2}dx$
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improper-integrals
1 Answers
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It's easier to see that
$$\frac{\partial^2I}{\partial a^2}=\int_0^\infty e^{-ax}\sin^2(bx)\ dx$$
Which after integration by parts:
$$\frac{\partial^2I}{\partial a^2}=\frac{2b^2}{a^3+4ab^2}=\frac1{2a}-\frac a{2(a^2+4b^2)}$$
And then integral with respect to $a$ twice and let $b=0$ and $a\to\infty$ to recover the constants of integration.
Of course, I assume that $a>0$ and $b\in[0,\pi)$.