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How to show that,

$$\int_{-\infty}^{+\infty}\frac{(x-1)^2}{(2x-1)^2+\left(ax^2-x-\frac{(a-1)^2}{2(2a-1)}\right)^2}\mathrm dx=\pi$$

Let assume $a\ge1$

I have tried substitution but it leads to a more complicated integral.

coolness
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  • Are you familiar with the Residue Theorem? – Franklin Pezzuti Dyer Jun 03 '19 at 20:51
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    I doubt residue theorem will work. For this integral someone got bored, took $\int_{-\infty}^\infty \frac{dx}{x^2+1}=\pi$ and applied Glasser's theorem. See: https://math.stackexchange.com/a/2821162/515527. Now who finds first what substitution was used (to undo the work), wins :) – Zacky Jun 03 '19 at 22:05

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Noting that \begin{eqnarray} &&(2x-1)^2+\left(ax^2-x-\frac{(a-1)^2}{2(2a-1)}\right)^2\\ &=&\left(ax^2-x-\frac{(a-1)^2}{2(2a-1)}+(2x-1)i\right)\left(ax^2-x-\frac{(a-1)^2}{2(2a-1)}-(2x-1)i\right) \\ &=&\left(ax^2+(2i-1)x-\frac{(a-1)^2}{2(2a-1)}-i\right)\left(ax^2-(2i+1)x-\frac{(a-1)^2}{2(2a-1)}+i\right) \end{eqnarray} so the equation $$(2x-1)^2+\left(ax^2-x-\frac{(a-1)^2}{2(2a-1)}\right)^2=0$$ has two roots in the upper half plane $$ x_1=\frac{(1+2 i)-\sqrt{\frac{2 a^3-(4+8 i) a^2-(4-12 i) a+(3-4 i)}{2 a-1}}}{2 a},x_2=\frac{(1+2 i)+\sqrt{\frac{2 a^3-(4+8 i) a^2-(4-12 i) a+(3-4 i)}{2 a-1}}}{2 a}.$$ Thus after long calculation, one has \begin{eqnarray} &&\int_{-\infty}^{+\infty}\frac{(x-1)^2}{(2x-1)^2+\left(ax^2-x-\frac{(a-1)^2}{2(2a-1)}\right)^2}\mathrm dx\\ &=&2\pi i\bigg[\text{Res}_{x=x_1}\bigg(\frac{(x-1)^2}{(2x-1)^2+\left(ax^2-x-\frac{(a-1)^2}{2(2a-1)}\right)^2}\bigg)\\ &&+\text{Res}_{x=x_2}\bigg(\frac{(x-1)^2}{(2x-1)^2+\left(ax^2-x-\frac{(a-1)^2}{2(2a-1)}\right)^2}\bigg)\bigg]\\ &=&2\pi i\bigg[\frac{(x_1^2-1)^2}{4(2x_1-1)+2\left(ax_1^2-x_1-\frac{(a-1)^2}{2(2a-1)}\right)(2ax_1-1)}\\ &&+\frac{(x_2^2-1)^2}{4(2x_2-1)+2\left(ax_2^2-x_2-\frac{(a-1)^2}{2(2a-1)}\right)(2ax_2-1)}\bigg]\\ &=&2\pi i(-\frac{i}{2})\\ &=&\pi. \end{eqnarray}

xpaul
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