$$\int_0^{\infty}\frac{\cos(z)\exp(-w z^2)-1}{z^2}dz=-\sqrt{\pi w}\,\exp\left(-\frac{1}{4w}\right)-\frac{\pi}{2}\,\operatorname{erf}\left(\frac{1}{2\sqrt{w}}\right)$$where w is positive and erf is the error function.
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3How do you have a $z$ in the final expression? I guess it should be some constant, perhaps, $2$? – sudeep5221 Jul 13 '19 at 14:32
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1Call it $f(w)$ and consider $f'(w)$. – metamorphy Jul 13 '19 at 14:52
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2It should be $-\sqrt{\pi w}$ at the beginning, not $-\sqrt{\pi z}$. – Varun Vejalla Jul 13 '19 at 15:35
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Thanks for pointing out my typo - corrected already – Honza Jul 13 '19 at 18:43
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Lt us write $$2I(w)=\int_{-\infty}^{\infty} \frac{\cos z~ e^{-w z^2}-1}{z^2} dz.$$ Differentiate w.r. $w$, then $$-2I'(w)=\int_{-\infty}^{\infty} \cos z ~ e^{-w z^2} dz= \Re \int_{-\infty}^{\infty} e^{-((wz^2-iz)} dz= e^{-1/(4w)}\int_{-\infty}^{\infty} e^{-wt^2} dt = \sqrt{\frac{\pi}{w} }\exp[-\frac{1}{4w}].$$ Integrating by letting $w=1/u^2$, we get $$-2I(w)=2\sqrt{\pi w} \exp[-\frac{1}{4w}]+\pi~ \mbox{erf}[\frac{1}{2\sqrt{w}}] +C$$ Note that $$2I(0)=\int_{-\infty}^{\infty} \frac{\cos z-1}{z^2}dz=-\pi \Rightarrow C=0.$$ Hence $$I(w)=-\sqrt{\pi w}\exp[\frac{-1}{4w}]-\frac{\pi}{2} \mbox{erf}[-\frac{1}{2\sqrt{w}}].$$
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