7.8-bo'lim : Noto'g'ri integrallar
8. Quyidagi integral yaqinlashish yoki ayrilish kerakligini aniqlang. Agar integral yaqinlashsa, uning qiymatini aniqlang.
\[\int_{{ - \infty }}^{\infty }{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}}\]Yechim
Birinchidan, chegaralardan birida cheksizlik mavjud ekan, biz integralni bajara olmasligimizni esga olishimiz kerak. Shuni ham yodda tutingki, bu holda bizda ikkala chegarada ham cheksizliklar mavjud va shuning uchun biz integralni ajratishimiz kerak.
Bu holda integral istalgan nuqtada bo'linishi mumkin va \(w = 0\) bo'linish nuqtasi uchun foydalanish uchun yaxshi nuqta kabi ko'rinadi. Integralni bo'lish,
\[\int_{{ - \infty }}^{\infty }{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}} = \int_{{ - \infty }}^{0}{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}} + \int_{0}^{\infty }{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}}\]Shuni ham yodda tutingki, har ikkala yangi integral konvergent bo'lsa, biz faqat shunday integralni buzishimiz mumkin! Agar ulardan bittasi ham divergent ekanligi aniqlansa, biz buni qila olmaganmiz va asl integral divergent bo'ladi.
Shunday qilib, buni amalga oshirish mumkinmi yoki yo'qmi, deb tashvishlanmang, keling, muammoni hal qilaylik.
Endi biz cheksizliklarni quyidagicha yo'q qilishimiz mumkin,
\[\int_{{ - \infty }}^{\infty }{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}} = \mathop {\lim }\limits_{t \to \,\, - \infty } \int_{t}^{0}{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}} + \mathop {\lim }\limits_{s \to \infty } \int_{0}^{s}{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}}\]2-qadam
Keyinchalik, integralni bajaramiz. Biz integratsiya jarayoniga ko'p tushuntirishlar/tafsilotlar kiritmaymiz. Shu nuqtaga kelib, sizning integratsiya qobiliyatingiz juda yaxshi bo'ladi deb taxmin qilinadi. Agar integratsiya qobiliyatingiz biroz zanglagan deb topsangiz, orqaga qaytib, oldingi tegishli bo'limlardan ba'zi mashqlarni bajarishingiz kerak.
Bunday holda biz oddiy Calc I almashtirishni amalga oshirishimiz mumkin. Mana integratsiya ishi.
\[\int{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}} = - \frac{3}{2}\frac{1}{{{w^4} + 1}} + c\]E'tibor bering, biz bu erda aniq integral qilmadik. Limitlar bizning integralni qanday bajarishimizga ta'sir qilmaydi va har biri uchun integral bir xil edi, faqat chegaralar boshqacha edi, shuning uchun integralni ikki marta bajarish uchun hech qanday sabab yo'q.
3-qadam
OK, endi integralning chegaralari haqida qayg'uraylik.
\[\begin{align*}\int_{{ - \infty }}^{\infty }{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}} & = \mathop {\lim }\limits_{t \to \,\, - \infty } \left. {\left( { - \frac{3}{2}\frac{1}{{{w^4} + 1}}} \right)} \right|_t^0 + \mathop {\lim }\limits_{s \to \infty } \left. {\left( { - \frac{3}{2}\frac{1}{{{w^4} + 1}}} \right)} \right|_0^s\\ & = \mathop {\lim }\limits_{t \to \,\, - \infty } \left( { - \frac{3}{2} + \frac{3}{2}\frac{1}{{{t^4} + 1}}} \right) + \mathop {\lim }\limits_{s \to \infty } \left( { - \frac{3}{2}\frac{1}{{{s^4} + 1}} + \frac{3}{2}} \right)\end{align*}\]4-qadam
Endi biz oldingi bosqichdagi javobimizdagi chegaralarni baholashimiz kerak. Mana ish chegarasi
\[\begin{align*}\int_{{ - \infty }}^{\infty }{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}} & = \mathop {\lim }\limits_{t \to \,\, - \infty } \left( { - \frac{3}{2} + \frac{3}{2}\frac{1}{{{t^4} + 1}}} \right) + \mathop {\lim }\limits_{s \to \infty } \left( { - \frac{3}{2}\frac{1}{{{s^4} + 1}} + \frac{3}{2}} \right)\\ & = \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ { - \frac{3}{2}} \right]\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\frac{3}{2}} \right]\end{align*}\]E'tibor bering, biz har bir chegara nima ekanligini tushunish uchun qavs ichiga javoblarni joylashtiramiz. Bu keyingi qadam uchun muhim bo'ladi.
5-qadam
Yakuniy qadam - javobni yozish!
Endi oldingi bosqichdagi ish chegarasidan shuni ko'ramizki,
\[\begin{align*}\int_{{ - \infty }}^{0}{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}} & = \mathop {\lim }\limits_{t \to \,\, - \infty } \left( { - \frac{3}{2} + \frac{3}{2}\frac{1}{{{t^4} + 1}}} \right) = \, - \frac{3}{2}\\ & \\ \int_{0}^{\infty }{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}} & = \mathop {\lim }\limits_{s \to \infty } \left( { - \frac{3}{2}\frac{1}{{{s^4} + 1}} + \frac{3}{2}} \right) = \frac{3}{2}\end{align*}\]Shuning uchun integrallarning har biri yaqinlashuvchi va yuqorida ko'rsatilgan qiymatlarga ega. Bu shuni anglatadiki, biz 1-bosqichda bo'lgani kabi integralni ham parchalashimiz mumkin. Bundan tashqari, asl integral endi,
\[\begin{align*}\int_{{ - \infty }}^{\infty }{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}} & = \int_{{ - \infty }}^{0}{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}} + \int_{0}^{\infty }{{\frac{{6{w^3}}}{{{{\left( {{w^4} + 1} \right)}^2}}}\,dw}}\\ & = \,\,\,\,\,\,\,\,\,\,\,\,\,\, - \frac{3}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{3}{2}\,\\ & = 0\end{align*}\]Shuning uchun integral birlashadi va uning qiymati 0.