Question Number 225652 by fantastic last updated on 05/Nov/25
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{{i}\pi{x}} \:{dx} \\ $$
Answered by mahdipoor last updated on 05/Nov/25
![(d/dx)(ae^(iπx) )=(iπa)e^(iπx) =e^(iπx) ⇒ iπa=1 ⇒a=(1/(πi))=−(i/π) ⇒∫_0 ^( π/2) e^(iπx) dx=[−(i/π)e^(iπx) ]_0 ^(π/2) = ((−i)/π)((cos(π^2 /2)+i.sin(π^2 /2))−1)](https://www.tinkutara.com/question/Q225653.png)
$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ae}^{\mathrm{i}\pi\mathrm{x}} \right)=\left(\mathrm{i}\pi\mathrm{a}\right)\mathrm{e}^{\mathrm{i}\pi\mathrm{x}} =\mathrm{e}^{\mathrm{i}\pi\mathrm{x}} \:\:\Rightarrow\:\mathrm{i}\pi\mathrm{a}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{a}=\frac{\mathrm{1}}{\pi\mathrm{i}}=−\frac{\mathrm{i}}{\pi} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\:\pi/\mathrm{2}} \mathrm{e}^{\mathrm{i}\pi\mathrm{x}} \mathrm{dx}=\left[−\frac{\mathrm{i}}{\pi}\mathrm{e}^{\mathrm{i}\pi\mathrm{x}} \right]_{\mathrm{0}} ^{\pi/\mathrm{2}} = \\ $$$$\frac{−\mathrm{i}}{\pi}\left(\left(\mathrm{cos}\left(\pi^{\mathrm{2}} /\mathrm{2}\right)+\mathrm{i}.\mathrm{sin}\left(\pi^{\mathrm{2}} /\mathrm{2}\right)\right)−\mathrm{1}\right) \\ $$
Answered by Ghisom_ last updated on 05/Nov/25
![∫_0 ^(π/2) e^(iπx) dx=∫_0 ^(π/2) (cos πx +i sin πx)dx= =(1/π)[sin πx −i cos πx]_0 ^(π/2) = =(1/π)(sin (π^2 /2)+(1−cos (π^2 /2))i) or ∫_0 ^(π/2) e^(iπx) dx=−(i/π)[e^(iπx) ]_0 ^(π/2) =(i/π)(1−e^(iπ^2 /2) ) both can be expressed as ((√(2−2cos (π^2 /2)))/π)e^(iπ^2 /4) ≈−.310469+.248096i](https://www.tinkutara.com/question/Q225665.png)
$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{e}^{\mathrm{i}\pi{x}} {dx}=\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\left(\mathrm{cos}\:\pi{x}\:+\mathrm{i}\:\mathrm{sin}\:\pi{x}\right){dx}= \\ $$$$=\frac{\mathrm{1}}{\pi}\left[\mathrm{sin}\:\pi{x}\:−\mathrm{i}\:\mathrm{cos}\:\pi{x}\right]_{\mathrm{0}} ^{\pi/\mathrm{2}} = \\ $$$$=\frac{\mathrm{1}}{\pi}\left(\mathrm{sin}\:\frac{\pi^{\mathrm{2}} }{\mathrm{2}}+\left(\mathrm{1}−\mathrm{cos}\:\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\right)\mathrm{i}\right) \\ $$$$\mathrm{or} \\ $$$$\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\mathrm{e}^{\mathrm{i}\pi{x}} {dx}=−\frac{\mathrm{i}}{\pi}\left[\mathrm{e}^{\mathrm{i}\pi{x}} \right]_{\mathrm{0}} ^{\pi/\mathrm{2}} =\frac{\mathrm{i}}{\pi}\left(\mathrm{1}−\mathrm{e}^{\mathrm{i}\pi^{\mathrm{2}} /\mathrm{2}} \right) \\ $$$$\mathrm{both}\:\mathrm{can}\:\mathrm{be}\:\mathrm{expressed}\:\mathrm{as} \\ $$$$\frac{\sqrt{\mathrm{2}−\mathrm{2cos}\:\frac{\pi^{\mathrm{2}} }{\mathrm{2}}}}{\pi}\mathrm{e}^{\mathrm{i}\pi^{\mathrm{2}} /\mathrm{4}} \approx−.\mathrm{310469}+.\mathrm{248096i} \\ $$