Evaluate the improper integral ${\int_0^{ + \infty } {\frac{{\cos t^2 - \sin t^2 }}{{1 + t^4 }}} dt}.$

Question

$$ \displaystyle{\int_0^{ + \infty } {\frac{{\cos t^2 - \sin t^2 }}{{1 + t^4 }}} dt}.$$

What I've done so far

$$\displaystyle \int_{0}^{+\infty}{\frac{\cos t^2-\sin t^2}{t^4+1}}dt=\frac{1}{2}\int_{ -\infty}^{+\infty}{\frac{\cos t^2-\sin t^2}{t^4+1}}dt.$$

But $\displaystyle \cos t^2 - \sin t^2 = \cos t^2 + \cos \left(\frac{\pi}{4} + t^2 \right) = \sqrt{2}\cos \left(\frac{\pi}{4} + t^2 \right)$

so $$\displaystyle \int_{-\infty}^{+\infty}{\frac{\cos t^2-\sin t^2}{t^4+1}}dt=\sqrt{2}\int_{- \infty}^{+\infty}{\frac{\cos \left(t^2+\frac{\pi}{4} \right)}{t^4+1}}dt.$$

Answer 1

Hint The form of the integral over $(-\infty, \infty)$ suggests using residue calculus. Here's an outline; I'll let you fill in the details.

1. Rewrite the original integral in exponential form (using evenness of the integrand as you have) as $$\int_0^\infty \frac{\cos t^2 - \sin t^2}{1 + t^4} \,dt = \frac{1}{2}\Re\left[(1 + i) \int_{-\infty}^\infty \frac{e^{i t^2} \,dt}{1 + t^4}\right] .$$
2. Form the contour integral of the integrand, regarded as a function of a complex variable $z$, over the boundary $\Gamma_R$ of a half-disk of radius $R$ centered at the origin in the upper half-plane. A standard estimate gives that the integral along the semicircular part of the boundary tends to $0$ as $R \to \infty$, so we have $$\int_{-\infty}^\infty \frac{e^{i t^2} \,dt}{1 + t^4} = \int_{\Gamma_R} \frac{e^{i z^2} \,dz}{1 + z^4} .$$
3. The integrand of the right-hand side is meromorphic, and its poles inside $\Gamma_R$ (for $R > 1$) are at $e^{\pi i / 4}, e^{3 \pi i / 4}$, so the Residue Theorem gives that $$\int_{-\infty}^\infty \frac{e^{i t^2} \,dt}{1 + t^4} = \int_{\Gamma_R} \frac{e^{i z^2} \,dz}{1 + z^4} = 2 \pi i \left[\operatorname{Res}\left(\frac{e^{i z^2}}{1 + z^4}; e^{\pi i / 4} \right) + \operatorname{Res}\left(\frac{e^{i z^2}}{1 + z^4}; e^{3 \pi / 4} \right)\right] . $$
4. Both poles are simple, making both residues straightforward to compute, yielding $$\int_{-\infty}^\infty \frac{e^{i t^2} \,dt}{1 + t^4} = \frac{\pi (1 - i)}{2 \sqrt 2} \left(\frac{1}{e} + i e\right) .$$
5. Substituting yields: \begin{multline*}\frac12 \Re\left[(1 + i) \int_{-\infty}^\infty \frac{e^{i t^2} \,dt}{1 + t^4}\right] \\ = \frac12 \Re \left[(1 + i) \cdot \frac{\pi (1 - i)}{2 \sqrt 2} \left(\frac{1}{e} + i e\right)\right] = \frac 12 \Re \left[\frac{\pi}{\sqrt 2}\left(\frac{1}{e} + i e\right)\right] = \boxed{\frac{\pi}{2 \sqrt 2 e}}.\end{multline*}

Remark Instead taking the imaginary part of the quantity in brackets gives the value of a related integral: $$\int_0^\infty \frac{\cos t^2 + \sin t^2}{1 + t^4} \,dt = \frac{\pi e}{2 \sqrt 2} .$$

Answer 2

We complete it $\displaystyle \frac{e^{iz^2}}{1 + z^4}$ in the first quadrant (positive real axis, quarter circle and positive imaginary axis).

We set $\displaystyle I = \int_0^{+ \infty} \frac{\cos t^2 - \sin t^2}{1 + t^4}$ dt.

From the real axis we have $$\displaystyle \int_0^{+ \infty} \frac{\cos t^2 + i \sin t^2}{1 + t^4} dt$$ while from the imaginary one $$\displaystyle -i \int_0^{+ \infty} \frac{\cos t^2 - i \sin t^2}{1 + t^4} dt$$, that is, in total $(1 - i)$ I. The integral over the quadrant tends to $0$.

The pole at $e^{i \pi / 4}$ has remainder $\displaystyle r = \frac{e^{-3 \pi i / 4}}{4 e}$, so the integral is equal to $\displaystyle 2 \pi i r = (1-i) \frac{\pi}{2 \sqrt{2} e}$.

Aha $\displaystyle I = \frac{\pi}{2 \sqrt{2} e}$

Answer 3

By differential equation

Let’s consider the parametrised integral $$ I(a)=\int_0^{\infty} \frac{\cos \left(a t^2\right)-\sin \left(a t^2\right)}{1+t^4} d t. $$ Differentiating $I(a)$ once and twice yields $$ I^{\prime}(a)=\int_0^{\infty} \frac{-t^2 \sin \left(a t^2\right)-t^2 \cos \left(a t^2\right)}{1+t^4} d t $$ and $$\begin{aligned} I^{\prime \prime}(a)&=\int_0^{\infty} \frac{-t^4 \cos \left(a t^2\right)+t^4 \sin \left(a t^2\right)}{1+t^4} d t\\&= I(a)+\int_0^{\infty}\left[\sin \left(a t^2\right)-\cos \left(a t^2\right)\right] d t\\&=I(a) \end{aligned} $$ Hence $$ I(a)=c_1 e^a+c_2 e^{-a} \tag*{(*)} $$ for some constants $c_1$ and $c_2$.

Putting the following results in $(*)$ $$ \begin{aligned} I(0) & =\int_0^{\infty} \frac{d t}{1+t^4} =\frac{\pi}{2 \sqrt{2}}; \end{aligned} $$ $$ \begin{aligned} I^{\prime}(0) & =-\int_0^{\infty} \frac{t^2}{1+t^4} d t=-\frac{\pi}{2 \sqrt{2}}, \end{aligned} $$ we have$$ I(a)=\frac{\pi e^{-a}}{2 \sqrt{2}} $$ In particular, setting $a=1$ yields our integral $$ I=\frac{\pi }{2 \sqrt{2}e} $$