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The following is an answer to your second question.

Using the approach Po1ynomial used here, we have $$ \begin{align} I = \int_{0}^{\infty} \frac{\sqrt{\sqrt{y^2+1}+1}}{\sqrt{y^2+1}} \cos(y) \,\mathrm{d}y &= \sqrt{2} \, \Re\int_{0}^{\infty} \frac{\cos (y)}{\sqrt{1+iy}} \, \mathrm dy \\ &= \frac{\sqrt{2}}{2} \, \Re \, \left(\int_{0}^{\infty}\frac{e^{iy}}{\sqrt{1+iy}} \, \mathrm dy + \int_{0}^{\infty} \frac{e^{-iy}}{\sqrt{1+iy}} \, \mathrm dy\right) . \end{align} $$

Integrating the integrand of the first integral around a quarter-circle contour in the first quadrant, and integrating the integrand of second integral around a quarter circle contour in the fourth quadrant, we get

$$ \begin{align} I &= \frac{\sqrt{2}}{2} \, \Re \, \left(i \int_{0}^{\infty} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm dy - i \int_{0}^{\infty} \frac{e^{-y}}{\sqrt{1+y}} \, \mathrm dy \right) \\ &=\frac{\sqrt{2}}{2} \, \Re \, \left(i \int_{0}^{1} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm dy + i \int_{1}^{\infty} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm dy- i \int_{0}^{\infty} \frac{e^{-y}}{\sqrt{1+y}} \, \mathrm dy \right) \\ &= \frac{\sqrt{2}}{2} \, \Re \, \left(i \int_{0}^{1} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm dy + \int_{1}^{\infty} \frac{e^{-y}}{\sqrt{y-1}} \, \mathrm dy- i \int_{0}^{\infty} \frac{e^{-y}}{\sqrt{1+y}} \, \mathrm dy \right) \\ &= \frac{\sqrt{2}}{2} \, \Re \, \left(i\int_{0}^{1} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm du + \frac{1}{e}\int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} \, \mathrm du -i \int_{0}^{\infty} \frac{e^{-y}}{\sqrt{1+y}} \, \mathrm dy \right) \\ &= \frac{\sqrt{2}}{2} \, \Re \, \left(\color{red}{i}\int_{0}^{1} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm dy + \frac{2}{e} \int_{0}^{\infty} e^{-w^{2}} \, \mathrm dw -\color{red}{i} \int_{0}^{\infty} \frac{e^{-y}}{\sqrt{1+y}} \, \mathrm dy \right) \\ &= \frac{\sqrt{2}}{e} \int_{0}^{\infty} e^{-w^{2}} \, \mathrm dw \\ &= \frac{\sqrt{2 \pi }}{2e}. \end{align}$$

Therefore, using your substitution, we have $$\int_{0}^{\infty} \cos(x\sqrt{x^2+2}) \,\mathrm{d}x = \frac{I}{2} = \frac{\sqrt{2 \pi}}{4e} = \frac{1}{e} \sqrt{\frac{\pi}{8}}.$$

The following is an answer to your second question.

Using the approach Po1ynomial used here, we have $$ \begin{align} I = \int_{0}^{\infty} \frac{\sqrt{\sqrt{y^2+1}+1}}{\sqrt{y^2+1}} \, \cos(y) \,\mathrm{d}y &= \sqrt{2} \, \Re\int_{0}^{\infty} \frac{\cos (y)}{\sqrt{1+iy}} \, \mathrm dy \\ &= \frac{\sqrt{2}}{2} \, \Re \, \left(\int_{0}^{\infty}\frac{e^{iy}}{\sqrt{1+iy}} \, \mathrm dy + \int_{0}^{\infty} \frac{e^{-iy}}{\sqrt{1+iy}} \, \mathrm dy\right) , \end{align} $$ where the square root is the principal branch of the square root.

Integrating the integrand of the first integral around a quarter-circle contour in the first quadrant, while integrating the integrand of second integral around a quarter circle contour in the fourth quadrant, we get

$$ \begin{align} I &= \frac{\sqrt{2}}{2} \, \Re \, \left(\int_{0}^{1} \frac{e^{-t}}{\sqrt{1-t}} \, \mathrm \, i \, dt + \int_{1}^{\infty} \frac{e^{-t}}{i \sqrt{t-1}} \, i \, \mathrm dt+ \int_{0}^{\infty} \frac{e^{-t}}{\sqrt{1+t}} \, (-i) \, \mathrm dt \right) \\ &= \frac{\sqrt{2}}{2} \, \Re \, \left(i\int_{0}^{1} \frac{e^{-t}}{\sqrt{1-t}} \, \mathrm dt + \frac{1}{e}\int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} \, \mathrm du - i \int_{0}^{\infty} \frac{e^{-t}}{\sqrt{1+t}} \, \mathrm dt \right) \\ &= \frac{\sqrt{2}}{2} \, \Re \, \left(\color{red}{i}\int_{0}^{1} \frac{e^{-t}}{\sqrt{1-t}} \, \mathrm dt + \frac{2}{e} \int_{0}^{\infty} e^{-w^{2}} \, \mathrm dw -\color{red}{i} \int_{0}^{\infty} \frac{e^{-t}}{\sqrt{1+t}} \, \mathrm dt \right) \\ &= \frac{\sqrt{2}}{e} \int_{0}^{\infty} e^{-w^{2}} \, \mathrm dw \\ &= \frac{\sqrt{2 \pi }}{2e}. \end{align}$$

Therefore, using your substitution, we have $$\int_{0}^{\infty} \cos(x\sqrt{x^2+2}) \,\mathrm{d}x = \frac{I}{2} = \frac{\sqrt{2 \pi}}{4e} = \frac{1}{e} \sqrt{\frac{\pi}{8}}.$$

The following is an answer to your second question.

Using the approach Po1ynomial used here, we have $$ \begin{align} I = \int_{0}^{\infty} \frac{\sqrt{\sqrt{y^2+1}+1}}{\sqrt{y^2+1}} \cos(y) \,\mathrm{d}y &= \sqrt{2} \, \Re\int_{0}^{\infty} \frac{\cos (y)}{\sqrt{1+iy}} \, \mathrm dy \\ &= \frac{\sqrt{2}}{2} \, \Re \, \left(\int_{0}^{\infty}\frac{e^{iy}}{\sqrt{1+iy}} \, \mathrm dy + \int_{0}^{\infty} \frac{e^{-iy}}{\sqrt{1+iy}} \, \mathrm dy\right) . \end{align} $$

Integrating the integrand of the first integral around a quarter-circle contour in the first quadrant, and integrating the integrand of second integral around a quarter circle contour in the fourth quadrant, we get

$$ \begin{align} I &= \frac{\sqrt{2}}{2} \, \Re \, \left(i \int_{0}^{\infty} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm dy - i \int_{0}^{\infty} \frac{e^{-y}}{\sqrt{1+y}} \, \mathrm dy \right) \\ &=\frac{\sqrt{2}}{2} \, \Re \, \left(i \int_{0}^{1} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm dy + i \int_{1}^{\infty} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm dy- i \int_{0}^{\infty} \frac{e^{-y}}{\sqrt{1+y}} \, \mathrm dy \right) \\ &= \frac{\sqrt{2}}{2} \, \Re \, \left(i \int_{0}^{1} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm dy + \int_{1}^{\infty} \frac{e^{-y}}{\sqrt{y-1}} \, \mathrm dy- i \int_{0}^{\infty} \frac{e^{-y}}{\sqrt{1+y}} \, \mathrm dy \right) \\ &= \frac{\sqrt{2}}{2} \, \Re \, \left(i\int_{0}^{1} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm du + \frac{1}{e}\int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} \, \mathrm dy -ie \int_{1}^{\infty} \frac{e^{-v}}{\sqrt{v}} \, \mathrm dv\right) \\ &= \frac{\sqrt{2}}{2} \, \Re \, \left(i\int_{0}^{1} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm dy + \frac{2}{e} \int_{0}^{\infty} e^{-w^{2}} \, \mathrm dw -2ie \int_{1}^{\infty} e^{-w^{2}} \, \mathrm dw\right) \\ &= \frac{\sqrt{2}}{e} \int_{0}^{\infty} e^{-w^{2}} \, \mathrm dw \\ &= \frac{\sqrt{2 \pi }}{2e}. \end{align}$$

Therefore, using your substitution, we have $$\int_{0}^{\infty} \cos(x\sqrt{x^2+2}) \,\mathrm{d}x = \frac{I}{2} = \frac{\sqrt{2 \pi}}{4e} = \frac{1}{e} \sqrt{\frac{\pi}{8}}.$$

The following is an answer to your second question.

Using the approach Po1ynomial used here, we have $$ \begin{align} I = \int_{0}^{\infty} \frac{\sqrt{\sqrt{y^2+1}+1}}{\sqrt{y^2+1}} \cos(y) \,\mathrm{d}y &= \sqrt{2} \, \Re\int_{0}^{\infty} \frac{\cos (y)}{\sqrt{1+iy}} \, \mathrm dy \\ &= \frac{\sqrt{2}}{2} \, \Re \, \left(\int_{0}^{\infty}\frac{e^{iy}}{\sqrt{1+iy}} \, \mathrm dy + \int_{0}^{\infty} \frac{e^{-iy}}{\sqrt{1+iy}} \, \mathrm dy\right) . \end{align} $$

Integrating the integrand of the first integral around a quarter-circle contour in the first quadrant, and integrating the integrand of second integral around a quarter circle contour in the fourth quadrant, we get

$$ \begin{align} I &= \frac{\sqrt{2}}{2} \, \Re \, \left(i \int_{0}^{\infty} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm dy - i \int_{0}^{\infty} \frac{e^{-y}}{\sqrt{1+y}} \, \mathrm dy \right) \\ &=\frac{\sqrt{2}}{2} \, \Re \, \left(i \int_{0}^{1} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm dy + i \int_{1}^{\infty} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm dy- i \int_{0}^{\infty} \frac{e^{-y}}{\sqrt{1+y}} \, \mathrm dy \right) \\ &= \frac{\sqrt{2}}{2} \, \Re \, \left(i \int_{0}^{1} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm dy + \int_{1}^{\infty} \frac{e^{-y}}{\sqrt{y-1}} \, \mathrm dy- i \int_{0}^{\infty} \frac{e^{-y}}{\sqrt{1+y}} \, \mathrm dy \right) \\ &= \frac{\sqrt{2}}{2} \, \Re \, \left(i\int_{0}^{1} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm du + \frac{1}{e}\int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} \, \mathrm du -i \int_{0}^{\infty} \frac{e^{-y}}{\sqrt{1+y}} \, \mathrm dy \right) \\ &= \frac{\sqrt{2}}{2} \, \Re \, \left(\color{red}{i}\int_{0}^{1} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm dy + \frac{2}{e} \int_{0}^{\infty} e^{-w^{2}} \, \mathrm dw -\color{red}{i} \int_{0}^{\infty} \frac{e^{-y}}{\sqrt{1+y}} \, \mathrm dy \right) \\ &= \frac{\sqrt{2}}{e} \int_{0}^{\infty} e^{-w^{2}} \, \mathrm dw \\ &= \frac{\sqrt{2 \pi }}{2e}. \end{align}$$

Therefore, using your substitution, we have $$\int_{0}^{\infty} \cos(x\sqrt{x^2+2}) \,\mathrm{d}x = \frac{I}{2} = \frac{\sqrt{2 \pi}}{4e} = \frac{1}{e} \sqrt{\frac{\pi}{8}}.$$

The following is an answer to your second question.

Using the approach Po1ynomial used here, we have $$ \begin{align} I = \int_{0}^{\infty} \frac{\sqrt{\sqrt{y^2+1}+1}}{\sqrt{y^2+1}} \cos(y) \,\mathrm{d}y &= \sqrt{2} \, \Re\int_{0}^{\infty} \frac{\cos (y)}{\sqrt{1+iy}} \, \mathrm dy \\ &= \frac{\sqrt{2}}{2} \, \Re \, \left(\int_{0}^{\infty}\frac{e^{iy}}{\sqrt{1+iy}} \, \mathrm dy + \int_{0}^{\infty} \frac{e^{-iy}}{\sqrt{1+iy}} \, \mathrm dy\right) . \end{align} $$

Integrating the integrand of the first integral around a quarter-circle contour in the first quadrant, and integrating the integrand of second integral around a quarter circle contour in the fourth quadrant, we get

$$ \begin{align} I &= \frac{\sqrt{2}}{2} \, \Re \, \left(i \int_{0}^{\infty} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm dy - i \int_{0}^{\infty} \frac{e^{-y}}{\sqrt{1+y}} \, \mathrm dy \right) \\ &= \frac{\sqrt{2}}{2} \, \Re \, \left(i \int_{0}^{1} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm dy + \int_{1}^{\infty} \frac{e^{-y}}{\sqrt{y-1}} \, \mathrm dy- i \int_{0}^{\infty} \frac{e^{-y}}{\sqrt{1+y}} \, \mathrm dy \right) \\ &= \frac{\sqrt{2}}{2} \, \Re \, \left(i\int_{0}^{1} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm du + \frac{1}{e}\int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} \, \mathrm dy -ie \int_{1}^{\infty} \frac{e^{-v}}{\sqrt{v}} \, \mathrm dv\right) \\ &= \frac{\sqrt{2}}{2} \, \Re \, \left(i\int_{0}^{1} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm dy + \frac{2}{e} \int_{0}^{\infty} e^{-w^{2}} \, \mathrm dw -2ie \int_{1}^{\infty} e^{-w^{2}} \, \mathrm dw\right) \\ &= \frac{\sqrt{2}}{e} \int_{0}^{\infty} e^{-w^{2}} \, \mathrm dw \\ &= \frac{\sqrt{2 \pi }}{2e}. \end{align}$$

Therefore, using your substitution, we have $$\int_{0}^{\infty} \cos(x\sqrt{x^2+2}) \,\mathrm{d}x = \frac{I}{2} = \frac{\sqrt{2 \pi}}{4e} = \frac{1}{e} \sqrt{\frac{\pi}{8}}.$$

The following is an answer to your second question.

Using the approach Po1ynomial used here, we have $$ \begin{align} I = \int_{0}^{\infty} \frac{\sqrt{\sqrt{y^2+1}+1}}{\sqrt{y^2+1}} \cos(y) \,\mathrm{d}y &= \sqrt{2} \, \Re\int_{0}^{\infty} \frac{\cos (y)}{\sqrt{1+iy}} \, \mathrm dy \\ &= \frac{\sqrt{2}}{2} \, \Re \, \left(\int_{0}^{\infty}\frac{e^{iy}}{\sqrt{1+iy}} \, \mathrm dy + \int_{0}^{\infty} \frac{e^{-iy}}{\sqrt{1+iy}} \, \mathrm dy\right) . \end{align} $$

Integrating the integrand of the first integral around a quarter-circle contour in the first quadrant, and integrating the integrand of second integral around a quarter circle contour in the fourth quadrant, we get

$$ \begin{align} I &= \frac{\sqrt{2}}{2} \, \Re \, \left(i \int_{0}^{\infty} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm dy - i \int_{0}^{\infty} \frac{e^{-y}}{\sqrt{1+y}} \, \mathrm dy \right) \\ &=\frac{\sqrt{2}}{2} \, \Re \, \left(i \int_{0}^{1} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm dy + i \int_{1}^{\infty} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm dy- i \int_{0}^{\infty} \frac{e^{-y}}{\sqrt{1+y}} \, \mathrm dy \right) \\ &= \frac{\sqrt{2}}{2} \, \Re \, \left(i \int_{0}^{1} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm dy + \int_{1}^{\infty} \frac{e^{-y}}{\sqrt{y-1}} \, \mathrm dy- i \int_{0}^{\infty} \frac{e^{-y}}{\sqrt{1+y}} \, \mathrm dy \right) \\ &= \frac{\sqrt{2}}{2} \, \Re \, \left(i\int_{0}^{1} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm du + \frac{1}{e}\int_{0}^{\infty} \frac{e^{-u}}{\sqrt{u}} \, \mathrm dy -ie \int_{1}^{\infty} \frac{e^{-v}}{\sqrt{v}} \, \mathrm dv\right) \\ &= \frac{\sqrt{2}}{2} \, \Re \, \left(i\int_{0}^{1} \frac{e^{-y}}{\sqrt{1-y}} \, \mathrm dy + \frac{2}{e} \int_{0}^{\infty} e^{-w^{2}} \, \mathrm dw -2ie \int_{1}^{\infty} e^{-w^{2}} \, \mathrm dw\right) \\ &= \frac{\sqrt{2}}{e} \int_{0}^{\infty} e^{-w^{2}} \, \mathrm dw \\ &= \frac{\sqrt{2 \pi }}{2e}. \end{align}$$

Therefore, using your substitution, we have $$\int_{0}^{\infty} \cos(x\sqrt{x^2+2}) \,\mathrm{d}x = \frac{I}{2} = \frac{\sqrt{2 \pi}}{4e} = \frac{1}{e} \sqrt{\frac{\pi}{8}}.$$