Differentiate the integral!

Integrate $\frac{e^{iaz^2}}{z^4+1}$ around the quadrant contour made up of the four parts:

• $\gamma_1$ , the line segment from $\varepsilon$ to $R$ , where $0 < \varepsilon < 1 < R$ ,

• $\gamma_2$ , the quadrant of a circle $z = Re^{i\theta}$ for $0 \le \theta \le \tfrac12\pi$ ,

• $\gamma_3$ , the line segment from $\varepsilon i$ to $R i$ ,

• $\gamma_4$ , the quadrant of a circle $z = \varepsilon e^{i\theta}$ for $0 \le \theta \le \tfrac12\pi$ .

Then $\left(\int_{\gamma_1} + \int_{\gamma_2} - \int_{\gamma_3} - \int_{\gamma_4}\right)\frac{e^{iaz^2}}{z^4+1}\,dz \; =\; 2\pi i\mathrm{Res}_{z=\frac{1+i}{\sqrt{2}}}\frac{e^{iaz^2}}{z^4+1} \; = \; 2\pi i \frac{e^{-a}}{4\left(\frac{1+i}{\sqrt{2}}\right)^3} \; = \; \tfrac{1}{2\sqrt{2}}\pi e^{-a}(1-i)$ Now $\left(\int_{\gamma_1} - \int_{\gamma_3}\right)\frac{e^{iaz^2}}{z^4+1}\,dz \; = \; \int_\varepsilon^R \frac{e^{iax^2} - ie^{-iax^2}}{x^4+1}\,dx$ while $\begin{aligned} \left|\int_{\gamma_2} \frac{e^{iaz^2}}{z^4+1}\,dz\right| & \le \; \int_0^{\frac12\pi} \frac{e^{-aR^2\sin2\theta}}{R^4-1}R\,d\theta \; \le \; \frac{\pi R}{2(R^4-1)} \\ \left|\int_{\gamma_4} \frac{e^{iaz^2}}{z^4+1}\,dz\right| & \le \; \int_0^{\frac12\pi} \frac{e^{-a\varepsilon^2\sin2\theta}}{1 - \varepsilon^4}\varepsilon\,d\theta \; \le \; \frac{\pi \varepsilon}{2(1 - \varepsilon^4)} \end{aligned}$ and hence both of these integrals tend to $0$ as $R \to \infty$ and $\varepsilon \to 0$ respectively. Thus we deduce that $\int_0^\infty \frac{e^{iax^2} - ie^{-iax^2}}{x^4+1}\,dx \; = \; \tfrac{1}{2\sqrt{2}}\pi e^{-a}(1-i)$ Taking the real part of this equation gives $\int_0^\infty \frac{\cos(ax^2) - \sin(ax^2)}{x^4+1}\,dx \; = \; \tfrac{1}{2\sqrt{2}}\pi e^{-a}$ and doubling this gives $I(a) \; = \; \tfrac{1}{\sqrt{2}}\pi e^{-a}$ provided always that $a \ge 0$ . Thus $10^5 I(7) \; = \; \frac{10^5 \pi}{\sqrt{2}e^7} \; = \; \boxed{202.5692413}$

We need to perform the calculation this way. Any attempt to differentiate under the integral sign, obtaining the differential equation $I''(a) =I(a)$ and so deducing that $I(a) = I(0)e^{-a}$ , is going to be difficult to achieve, since validity of the likely "formula" for the second derivative $I''(a) \; = \; -\int_{\mathbb{R}} \frac{x^4(\cos(ax^2)-\sin(ax^2))}{x^4+1}\,dx$ is dependent on interchanging the order of limits (differentiation with respect to $a$ and integration with respect to $x$ ) where the $x$ integrals are at best conditionally convergent.

I randomly was pressing buttons and I got it!

I get answer as $I(a)=\frac {\pi}{\sqrt 2} \left( \cosh (a) -\sinh (a)\right)$

Using contour integration taking a semicircular contour.

@Richeek Das

You know what, I used Feynman's Trick to solve this and i ended up with two options, Either $y=\dfrac{\pi}{\sqrt{2}} e^{-x}$ or $y=\dfrac{\pi}{\sqrt{2}} e^{x}$

I used both and got the answer in the first expression. I dont know how to solve by this method so that i get only the req. answer.

So When i differentiated it $I(a)$ twice, I got $I(a)=I''(a)$ .

So this left me with multiple options like $ksin(a)$ or $ke^{a}$ or $ke^{-a}$

Last one worked .. I inserted a=0 to get k.

I know my method is poor :P