In mathematics, there are several integrals known as the Dirichlet integral, after the German mathematician Peter Gustav Lejeune Dirichlet, one of which is the improper integral of the sinc function over the positive real number line.
Dalam matematika, ada beberapa integral yang dikenal sebagai Integral Dirichlet, setelah ahli matematika Jerman Peter Gustav Lejeune Dirichlet, salah satunya adalah integral tak wajar dari fungsi sinc di atas garis nyata positif: ∫ 0 ∞ sin ⁡ x x d
∫ 0 ∞ sin ⁡ x x d xπ 2 .
x d x∫ 0 ∞ sin ⁡ ( x ) x d xπ 2 .
This integral is not absolutely convergent, meaning | sin ⁡ x x | {x}}\right|} has an infinite Lebesgue or Riemann improper integral over the positive real line, so the sinc function is not Lebesgue integrable over the positive real line.
^{\infty }{\frac {\sin x}{x}}\,dx={\frac {\pi }{2}}.} Integral ini bukanlah absolut konvergen, artinya | sin ⁡ x x | {\frac {\sin x}{x}}{\Biggl |}} bukan Lebesgue-integrable, sehingga integral Dirichlet tidak terdefinisi dalam arti integral Lebesgue.
Let f ( t ) be a function defined whenever t ≥ 0. Then its Laplace transform is given by L { f ( t ) }
Maka f ( t ) menjadi fungsi yang dapat didefinisikan t ≥ 0 .
[ f ( t ) t ]∫ s ∞ F ( u ) d u , }\left[{\frac {f(t)}{t}}\right]=\int _{s}^{\infty }F(u)\,du,} provided lim t → 0 f ( t ) t {\frac {f(t)}{t}}} exists.
d u , }{\Biggl [}{\frac {f(t)}{t}}{\Biggl ]}=\int _{s}^{\infty }F(u)\,du,} asalkan lim t → 0 f ( t ) t {\frac {f(t)}{t}}} .
Therefore, ∫ 0 ∞ sin ⁡ t t d tlim s → 0 ∫ 0 ∞ e − s t sin ⁡ t t d tlim s → 0 L [ sin ⁡ t t ]lim s → 0 ∫ s ∞ d u u 2 + 1lim s → 0 arctan ⁡ u | s ∞lim s → 0 [ π 2 − arctan ⁡ ( s ) ]π 2 .
Maka kami ingin mengatakan ∫ 0 ∞ sin ⁡ ( t ) t d tlim λ → ∞ ∫ 0 λ π 2 sin ⁡ ( t ) t d tlim λ → ∞ ∫ 0 π 2 sin ⁡ ( λ x )
( I 2∫ 0 ∞ ∫ 0 ∞ e − s t sin ⁡ t d s d t ) ,\int _{0}^{\infty }\int _{0}^{\infty }e^{-st}\sin t\,dt\,ds\right)=\left(I_{2}=\int _{0}^{\infty }\int _{0}^{\infty }e^{-st}\sin t\,ds\,dt\right),} ( I 1∫ 0 ∞ 1 s 2 + 1 d sπ 2 )( I 2∫ 0 ∞ sin ⁡ t t d t ) , provided s > 0.
={\frac {1}{2i}}\left(e^{i\omega }-e^{-i\omega }\right).} oleh karena itu, d f d a− ∫ 0 ∞ e − a ω sin ⁡ ω d ω− ∫ 0 ∞ e − a ω e i ω − e − i ω 2 i d ω− 1 2 i ∫ 0 ∞
Differentiate with respect to s > 0 and apply the Leibniz rule for differentiating under the integral sign to obtain d f d sd d s ∫ 0 ∞ e − s t sin ⁡ t t d t∫ 0 ∞ ∂ ∂ s e − s t sin ⁡ t t d t− ∫ 0 ∞ e − s t sin ⁡ t d t .
Diferensialkan sehubungan dengan a dan terapkan hukum Leibniz untuk membedakan di bawah tanda integral untuk mendapatkan d f d ad d a ∫ 0 ∞ e − a ω sin ⁡ ω ω d ω∫ 0 ∞ ∂ ∂ a e − a ω sin ⁡ ω ω d ω− ∫ 0 ∞ e − a ω sin ⁡ ω d ω .
{\frac {df}{ds}}&={\frac {d}{ds}}\int _{0}^{\infty }e^{-st}{\frac {\sin t}{t}}\,dt=\int _{0}^{\infty }{\frac {\partial }{\partial s}}e^{-st}{\frac {\sin t}{t}}\,dt\\[6pt]&=-\int _{0}^{\infty }e^{-st}\sin t\,dt.\end{aligned}}} Now, using Euler's formula e i tcos ⁡ t + i sin ⁡ t ,\cos t+i\sin t,} one can express the sine function in terms of complex exponentials: sin ⁡ t1 2 i ( e i t − e − i t ) .
{\frac {df}{da}}&={\frac {d}{da}}\int _{0}^{\infty }e^{-a\omega }{\frac {\sin \omega }{\omega }}\,d\omega\int _{0}^{\infty }{\frac {\partial }{\partial a}}e^{-a\omega }{\frac {\sin \omega }{\omega }}\,d\omega \\[6pt]&=-\int _{0}^{\infty }e^{-a\omega }\sin \omega \,d\omega .\end{aligned}}} Sekarang, gunakan rumus Euler e i ωcos ⁡ ω + i sin ⁡ ω ,\cos \omega +i\sin \omega ,} sinusoid dapat dinyatakan dalam fungsi eksponensial kompleks.
{\frac {df}{ds}}&=-\int _{0}^{\infty }e^{-st}\sin t\,dt=-\int _{0}^{\infty }e^{-st}{\frac {e^{it}-e^{-it}}{2i}}dt\\[6pt]&=-{\frac {1}{2i}}\int _{0}^{\infty }\left[e^{-t(s-i)}-e^{-t(s+i)}\right]dt\\[6pt]&=-{\frac {1}{2i}}\left[{\frac {-1}{s-i}}e^{-t(s-i)}-{\frac {-1}{s+i}}e^{-t(s+i)}\right]_{0}^{\infty }\\[6pt]&=-{\frac {1}{2i}}\left[0-\left({\frac {-1}{s-i}}+{\frac {1}{s+i}}\right)\right]=-{\frac {1}{2i}}\left({\frac {1}{s-i}}-{\frac {1}{s+i}}\right)\\[6pt]&=-{\frac {1}{2i}}\left({\frac {s+i-(s-i)}{s^{2}+1}}\right)=-{\frac {1}{s^{2}+1}}.\end{aligned}}} Integrating with respect to s gives f ( s )∫ − d s s 2 + 1A − arctan ⁡ s , {s^{2}+1}}=A-\arctan s,} where A is a constant of integration to be determined.
{\frac {df}{da}}&={\frac {d}{da}}\int _{0}^{\infty }e^{-a\omega }{\frac {\sin \omega }{\omega }}\,d\omega\int _{0}^{\infty }{\frac {\partial }{\partial a}}e^{-a\omega }{\frac {\sin \omega }{\omega }}\,d\omega \\[6pt]&=-\int _{0}^{\infty }e^{-a\omega }\sin \omega \,d\omega .\end{aligned}}} Sekarang, gunakan rumus Euler e i ωcos ⁡ ω + i sin ⁡ ω ,\cos \omega +i\sin \omega ,} sinusoid dapat dinyatakan dalam fungsi eksponensial kompleks.
Since lim s → ∞ f ( s )
Karena lim a → ∞ f ( a )
0 , f(s)=0,} Alim s → ∞ arctan ⁡ sπ 2 , \arctan s={\frac {\pi }{2}},} using the principal value.
0 , f(a)=0,} Alim a → ∞ arctan ⁡ ( a )π 2 , \arctan(a)={\frac {\pi }{2}},} menggunakan nilai pokok.
π 2 − arctan ⁡ ( 0 )π 2 , {2}}-\arctan(0)={\frac {\pi }{2}},} as before.
π 2 {2}}-\arctan(0)={\frac {\pi }{2}}} , seperti sebelumnya.
As a function of the complex variable z , it has a simple pole at the origin, which prevents the application of Jordan's lemma, whose other hypotheses are satisfied.
Sebagai fungsi dari variabel kompleks z , ia memiliki kutub sederhana di asalnya, yang mencegah penerapan lemma Jordan, yang hipotesis lainnya terpenuhi.
The pole has been moved to the negative imaginary axis, so g ( z ) can be integrated along the semicircle γ of radius R centered at z0 extending in the positive imaginary direction, and closed along the real axis.
Kutub telah dipindahkan dari sumbu sebenarnya, sehingga g ( z ) dapat diintegrasikan sepanjang setengah lingkaran radius R yang berpusat di z0 dan ditutup pada sumbu nyata.
One then takes the limit ε → 0.
Seseorang kemudian limitnya ε → 0 .
The complex integral is zero by the residue theorem, as there are no poles inside the integration path γ : 0∫ γ g ( z )
Integral kompleks adalah nol menurut teorema residu, karena tidak ada kutub di dalam jalur integrasi 0∫ γ g ( z )
The second term vanishes as R goes to infinity.
Istilah kedua lenyap saat R pergi ke tak terhingga.
As for the first integral, one can use one version of the Sokhotski–Plemelj theorem for integrals over the real line: for a complex-valued function f defined and continuously differentiable on the real line and real constants a and b with a < 0 < b one finds lim ε → 0 + ∫ a b f ( x )
Adapun integral pertama, seseorang dapat menggunakan satu versi teorema Sokhotski–Plemelj untuk integral di atas garis nyata: untuk fungsi bernilai kompleks f yang ditentukan dan dapat terus terdiferensiasi pada garis nyata dan konstanta nyata a dan b with a < 0 < b seseorang menemukan lim ε → 0 + ∫ a b f ( x ) x ± i ε d x∓ i π f ( 0 ) + P ∫ a b f ( x ) x d x , }\int _{a}^{b}{\frac {f(x)}{x\pm i\varepsilon }}\,dx=\mp i\pi f(0)+{\mathcal {P}}\int _{a}^{b}{\frac {f(x)}{x}}\,dx,} di mana P }} menunjukkan nilai pokok Cauchy.
Back to the above original calculation, one can write 0P ∫ e i x x d x − π i .
Kembali ke kalkulasi awal di atas, seseorang dapat menulis 0P ∫ e i x x d x − π i .
}\int {\frac {e^{ix}}{x}}\,dx-\pi i.} By taking the imaginary part on both sides and noting that the function sin ⁡ ( x ) / x is even, we get ∫ − ∞ + ∞ sin ⁡ ( x )
}\int {\frac {e^{ix}}{x}}\,dx-\pi i.} Dengan mengambil bagian imajiner di kedua sisi dan mencatat fungsinya sin ⁡ ( x ) / x bahkan, kita dapatkan ∫ − ∞ + ∞ sin ⁡ ( x ) x d x2 ∫ 0 + ∞ sin ⁡ ( x ) x d x .
x d x2 ∫ 0 + ∞ sin ⁡ ( x ) x d x .
x d x∫ 0 ∞ sin ⁡ ( x ) x d xπ 2 .
^{+\infty }{\frac {\sin(x)}{x}}\,dx=2\int _{0}^{+\infty }{\frac {\sin(x)}{x}}\,dx.}
^{+\infty }{\frac {\sin(x)}{x}}\,dx=2\int _{0}^{+\infty }{\frac {\sin(x)}{x}}\,dx.} Akhirnya, lim ε → 0 ∫ ε ∞ sin ⁡ ( x )
Finally, lim ε → 0 ∫ ε ∞ sin ⁡ ( x ) x d x∫ 0 ∞ sin ⁡ ( x ) x d xπ 2 .
x ) sin ⁡ ( x ) d xlim n → ∞ ∫ 0 π 2 D n ( x )
\int _{\varepsilon }^{\infty }{\frac {\sin(x)}{x}}\,dx=\int _{0}^{\infty }{\frac {\sin(x)}{x}}\,dx={\frac {\pi }{2}}.} Alternatively, choose as the integration contour for f the union of upper half-plane semicircles of radii ε and R together with two segments of the real line that connect them.
\int _{\varepsilon }^{\infty }{\frac {\sin(x)}{x}}\,dx=\int _{0}^{\infty }{\frac {\sin(x)}{x}}\,dx={\frac {\pi }{2}}.} Atau, pilih sebagai kontur integrasi untuk f gabungan jari-jari setengah lingkaran bidang atas ε dan R bersama dengan dua segmen dari garis nyata yang menghubungkannya.
On one hand the contour integral is zero, independently of ε and R ; on the other hand, as ε → 0 and R → ∞ the integral's imaginary part converges to 2 I + ℑ ( ln ⁡ 0 − ln ⁡ ( π i ) )
Di satu sisi, integral kontur adalah nol, terlepas dari ε dan R ; di sisi lain, sebagai ε → 0 dan R → ∞ bagian imajiner integral menyatu 2 I + ℑ ( ln ⁡ 0 − ln ⁡ ( π i ) )
2 I − π \ln 0-\ln(\pi i){\big )}=2I-\pi } (here ln ⁡ z is any branch of logarithm on upper half-plane), leading to Iπ 2 .
2 I − π \ln 0-\ln(\pi i){\big )}=2I-\pi } (maka ln ⁡ z adalah cabang dari logaritma pada setengah bidang atas), yang mengarah ke Iπ 2 {2}}} .
1 + 2 ∑ k1 n cos ⁡ ( 2 k x )sin ⁡ [ ( 2 n + 1 )
− 1 sin ⁡ ( x ) + x cos ⁡ ( x )lim x → 0 − sin ⁡ ( x ) 2 cos ⁡ ( x ) − x sin ⁡ ( x )
x ] sin ⁡ ( x ) .
− x x sin ⁡ ( x )
It immediately follows that: ∫ 0 π 2 D n ( x )
Segera setelah itu ∫ 0 π 2 D n ( x )
d xπ 2 . ^{\frac {\pi }{2}}D_{n}(x)\,dx={\frac {\pi }{2}}.} Define f ( x ){ 1 x − 1 sin ⁡ ( x ) x ≠ 0 0 x0 {\frac {1}{x}}-{\frac {1}{\sin(x)}}&x\neq 0\\[6pt]0&x=0\end{cases}}} Clearly, f is continuous when x ∈ ( 0 , π / 2 ] ; to see its continuity at 0 apply L'Hopital's Rule: lim x → 0 sin ⁡ ( x ) − x x sin ⁡ ( x )
d xπ 2 . ^{\frac {\pi }{2}}D_{n}(x)dx={\frac {\pi }{2}}.} menjelaskan f ( x ){ 1 x − 1 sin ⁡ ( x ) x ≠ 0 0 x0 {\frac {1}{x}}-{\frac {1}{\sin(x)}}&x\neq 0\\[6pt]0&x=0\end{cases}}} Jelasnya, f adalah kontinu jika x ≠ 0 , untuk melihat keberlanjutannya di 0 terapkan Aturan L'Hopital: lim x → 0 sin ⁡ ( x )
0. {\frac {\sin(x)-x}{x\sin(x)}}=\lim _{x\to 0}{\frac {\cos(x)-1}{\sin(x)+x\cos(x)}}=\lim _{x\to 0}{\frac {-\sin(x)}{2\cos(x)-x\sin(x)}}=0.} Hence, f fulfills the requirements of the Riemann-Lebesgue Lemma.
0. {\frac {\sin(x)-x}{x\sin(x)}}=\lim _{x\to 0}{\frac {\cos(x)-1}{\sin(x)+x\cos(x)}}=\lim _{x\to 0}{\frac {-\sin(x)}{2\cos(x)-x\sin(x)}}=0.} Karenanya, f memenuhi persyaratan Riemann-Lebesgue Lemma.
This means: lim λ → ∞ ∫ 0 π / 2 f ( x ) sin ⁡ ( λ x )
Ini berarti lim λ → ∞ ∫ a b f ( x ) sin ⁡ ( λ x )
d x0 ⟹ lim λ → ∞ ∫ 0 π / 2 sin ⁡ ( λ x ) x d xlim λ → ∞ ∫ 0 π / 2 sin ⁡ ( λ x ) sin ⁡ ( x )
x d xlim λ → ∞ ∫ 0 π 2 sin ⁡ ( λ x ) sin ⁡ ( x )
\int _{0}^{\pi /2}f(x)\sin(\lambda x)dx=0\quad \Longrightarrow \quad \lim _{\lambda \to \infty }\int _{0}^{\pi /2}{\frac {\sin(\lambda x)}{x}}dx=\lim _{\lambda \to \infty }\int _{0}^{\pi /2}{\frac {\sin(\lambda x)}{\sin(x)}}dx.}
\int _{a}^{b}f(x)\sin(\lambda x)dx=0\Rightarrow \lim _{\lambda \to \infty }\int _{a}^{b}{\frac {\sin(\lambda x)}{x}}dx=\lim _{\lambda \to \infty }\int _{a}^{b}{\frac {\sin(\lambda x)}{\sin(x)}}dx.} (Bentuk Lemma Riemann-Lebesgue yang digunakan di sini dibuktikan dalam artikel yang dikutip.)
We would like to compute: ∫ 0 ∞ sin ⁡ ( t ) t d tlim λ → ∞ ∫ 0 λ π 2 sin ⁡ ( t ) t d tlim λ → ∞ ∫ 0 π 2 sin ⁡ ( λ x )
Maka kami ingin mengatakan ∫ 0 ∞ sin ⁡ ( t ) t d tlim λ → ∞ ∫ 0 λ π 2 sin ⁡ ( t ) t d tlim λ → ∞ ∫ 0 π 2 sin ⁡ ( λ x )
d xπ 2 \int _{0}^{\infty }{\frac {\sin(t)}{t}}dt=&\lim _{\lambda \to \infty }\int _{0}^{\lambda {\frac {\pi }{2}}}{\frac {\sin(t)}{t}}dt\\[6pt]=&\lim _{\lambda \to \infty }\int _{0}^{\frac {\pi }{2}}{\frac {\sin(\lambda x)}{x}}dx\\[6pt]=&\lim _{\lambda \to \infty }\int _{0}^{\frac {\pi }{2}}{\frac {\sin(\lambda x)}{\sin(x)}}dx\\[6pt]=&\lim _{n\to \infty }\int _{0}^{\frac {\pi }{2}}{\frac {\sin((2n+1)x)}{\sin(x)}}dx\\[6pt]=&\lim _{n\to \infty }\int _{0}^{\frac {\pi }{2}}D_{n}(x)dx={\frac {\pi }{2}}\end{aligned}}} However, we must justify switching the real limit in λ to the integral limit in n , which will follow from showing that the limit does exist.
d xπ 2 \int _{0}^{\infty }{\frac {\sin(t)}{t}}dt=&\lim _{\lambda \to \infty }\int _{0}^{\lambda {\frac {\pi }{2}}}{\frac {\sin(t)}{t}}dt\\[6pt]=&\lim _{\lambda \to \infty }\int _{0}^{\frac {\pi }{2}}{\frac {\sin(\lambda x)}{x}}dx\\[6pt]=&\lim _{\lambda \to \infty }\int _{0}^{\frac {\pi }{2}}{\frac {\sin(\lambda x)}{\sin(x)}}dx\\[6pt]=&\lim _{n\to \infty }\int _{0}^{\frac {\pi }{2}}{\frac {\sin((2n+1)x)}{\sin(x)}}dx\\[6pt]=&\lim _{n\to \infty }\int _{0}^{\frac {\pi }{2}}D_{n}(x)dx={\frac {\pi }{2}}\end{aligned}}} In order to do so, however, we must justify switching the real limit in λ to the integral limit in n .
Using the Taylor-series expansion of the cosine about zero, 1 − cos ⁡ ( x )1 − ∑ k ≥ 0 ( − 1 ) ( k + 1 ) x 2 k 2 k !
Using the Taylor-series expansion of the cosine about zero, 1 − cos ⁡ ( x )1 − ∑ k ≥ 0 x 2 k 2 k !
∑ k ≥ 1 ( − 1 ) ( k + 1 ) x 2 k 2 k ! .
− ∑ k ≥ 1 x 2 k 2 k ! .
{\frac {{(-1)^{(k+1)}}x^{2k}}{2k!}}=\sum _{k\geq 1}{\frac {{(-1)^{(k+1)}}x^{2k}}{2k!}}.}
{\frac {x^{2k}}{2k!}}=-\sum _{k\geq 1}{\frac {x^{2k}}{2k!}}.}
x 2 | d x ≤ ∫ − ∞ − ε 2 x 2 d x + ∫ − ε ε e | x | d x + ∫ ε ∞ 2 x 2 d x ≤ K , ^{\infty }\left|{\frac {1-\cos(x)}{x^{2}}}\right|dx\leq \int _{-\infty }^{-\varepsilon }{\frac {2}{x^{2}}}dx+\int _{-\varepsilon }^{\varepsilon }e^{|x|}dx+\int _{\varepsilon }^{\infty }{\frac {2}{x^{2}}}dx\leq K,} for some constant K > 0.
x 2 | d x ≤ ∫ − ∞ − ε 2 x 2 d x + ∫ − ε ε e | x | d x + ∫ ε ∞ 2 x 2 d x ≤ K , ^{\infty }\left|{\frac {1-\cos(x)}{x^{2}}}\right|dx\leq \int _{-\infty }^{-\varepsilon }{\frac {2}{x^{2}}}dx+\int _{-\varepsilon }^{\varepsilon }e^{|x|}dx+\int _{\varepsilon }^{\infty }{\frac {2}{x^{2}}}dx\leq K,} for some constant K > 0 .
Dirichlet distribution Dirichlet principle Sinc function Fresnel integral Weisstein, Eric W.
Distribusi Dirichlet Prinsip Dirichlet Fungsi Sin Integral Fresnel (Inggris) Weisstein, Eric W.