∫sinx/xdx定积分0到1是可积的吗？ 求∫sinx/xdx
sinx/x\u57280\u5230\u221e\u7684\u5b9a\u79ef\u5206\uff0c\u5177\u4f53\u6b65\u9aa4\u5bf9sinx\u6cf0\u52d2\u5c55\u5f00,\u518d\u9664\u4ee5x\u6709\uff1a
sinx/x=1-x^2/3!+x^4/5!+\u2026+(-1)^(m-1)x^(2m-2)/(2m-1)!+o(1)
\u4e24\u8fb9\u6c42\u79ef\u5206\u6709\uff1a
\u222bsinx/x\u00b7dx
=[x/1-x^3/3\u00b73!+x^5/5\u00b75!+\u2026+(-1)^(m-1)x^(2m-1)/(2m-1)(2m-1)!+o(1)]
\u4ece0\u5230\u65e0\u7a77\u5b9a\u79ef\u5206
\u5219\u5c060,x(x\u219200\uff09\uff08\u8fd9\u91cc\u7684x\u662f\u4e00\u4e2a\u5f88\u5927\u7684\u5e38\u6570,\u53ef\u4ee5\u4efb\u610f\u53d6\uff09\u4ee3\u5165\u4e0a\u5f0f\u53f3\u8fb9\u5e76\u76f8\u51cf,\u901a\u8fc7\u8ba1\u7b97\u673a\u5373\u53ef\u5f97\u5230\u7ed3\u679c
\u4ee5\u4e0a\u53ea\u662f\u4e2a\u4eba\u610f\u89c1,\u4ee5\u4e0b\u662f\u9ad8\u624b\u7684\u505a\u6cd5\uff1a
\uff08\u9ad8\u624b\u51fa\u9a6c,\u975e\u540c\u51e1\u54cd!\uff09
\u8003\u8651\u5e7f\u4e49\u4e8c\u91cd\u79ef\u5206
I=\u222b\u222b e^(-xy) \u00b7sinxdxdy
D
\u5176\u4e2dD = [0,+\u221e)\u00d7[0,+\u221e),
\u4eca\u6309\u4e24\u79cd\u4e0d\u540c\u7684\u6b21\u5e8f\u8fdb\u884c\u79ef\u5206\u5f97
I=\u222bsinxdx \u222be^(-xy)dy
0 +\u221e 0 +\u221e
= \u222bsinx\u00b7(1/x)dx
0 +\u221e
\u53e6\u4e00\u65b9\u9762,\u4ea4\u6362\u79ef\u5206\u987a\u5e8f\u6709\uff1a
I=\u222b\u222b e^(-xy) \u00b7sinxdxdy
D
=\u222bdy \u222be^(-xy)\u00b7sinxdx
0 +\u221e 0 +\u221e
=\u222bdy/(1+y^2\uff09=arc tan+\u221e-arc tan0
0 +\u221e
= \u03c0/2
\u6240\u4ee5\uff1a
\u222bsinx\u00b7(1/x)dx=\u03c0/2
0 +\u221e
\u6269\u5c55\u8d44\u6599\u5e38\u7528\u79ef\u5206\u516c\u5f0f\uff1a
1\uff09\u222b0dx=c
2\uff09\u222bx^udx=(x^(u+1))/(u+1)+c
3\uff09\u222b1/xdx=ln|x|+c
4\uff09\u222ba^xdx=(a^x)/lna+c
5\uff09\u222be^xdx=e^x+c
6\uff09\u222bsinxdx=-cosx+c
7\uff09\u222bcosxdx=sinx+c
8\uff09\u222b1/(cosx)^2dx=tanx+c
9\uff09\u222b1/(sinx)^2dx=-cotx+c
10\uff09\u222b1/\u221a\uff081-x^2) dx=arcsinx+c
\u4e00\u822c\u5b9a\u7406
\u5b9a\u74061\uff1a\u8bbef(x)\u5728\u533a\u95f4[a,b]\u4e0a\u8fde\u7eed\uff0c\u5219f(x)\u5728[a,b]\u4e0a\u53ef\u79ef\u3002
\u5b9a\u74062\uff1a\u8bbef(x)\u533a\u95f4[a,b]\u4e0a\u6709\u754c\uff0c\u4e14\u53ea\u6709\u6709\u9650\u4e2a\u95f4\u65ad\u70b9\uff0c\u5219f(x)\u5728[a,b]\u4e0a\u53ef\u79ef\u3002
\u5b9a\u74063\uff1a\u8bbef(x)\u5728\u533a\u95f4[a,b]\u4e0a\u5355\u8c03\uff0c\u5219f(x)\u5728[a,b]\u4e0a\u53ef\u79ef\u3002
\u222bsinx/x dx\u4e0d\u80fd\u7528\u521d\u7b49\u51fd\u6570\u8868\u793a
I=\u222b\u222b{D}siny/y dxdy
=\u222b{0->1}dy \u222b{y^2->y}siny/ydx
=\u222b{0->1}(siny/y) (y-y^2)dy
=\u222b{0->1}(siny-siny*y)dy
=\u222b{0->1}(1-y)d[-cosy]
=(1-1)[-cos1]-(1-0)d[-cos0]
=\u222b{0->1}[-cosy]d[1-y]
=1-\u222b{0->1}cosydy
=1-sin1
\u79ef\u5206\u662f\u7ebf\u6027\u7684\uff0c\u5982\u679c\u4e00\u4e2a\u51fd\u6570f\u53ef\u79ef\uff0c\u90a3\u4e48\u5b83\u4e58\u4ee5\u4e00\u4e2a\u5e38\u6570\u540e\u4ecd\u7136\u53ef\u79ef\u3002\u5982\u679c\u51fd\u6570f\u548cg\u53ef\u79ef\uff0c\u90a3\u4e48\u5b83\u4eec\u7684\u548c\u4e0e\u5dee\u4e5f\u53ef\u79ef\u3002
\u6269\u5c55\u8d44\u6599\uff1a
\u5982\u679c\u4e00\u4e2a\u51fd\u6570f\u5728\u67d0\u4e2a\u533a\u95f4\u4e0a\u9ece\u66fc\u53ef\u79ef\uff0c\u5e76\u4e14\u5728\u6b64\u533a\u95f4\u4e0a\u5927\u4e8e\u7b49\u4e8e\u96f6\u3002\u90a3\u4e48\u5b83\u5728\u8fd9\u4e2a\u533a\u95f4\u4e0a\u7684\u79ef\u5206\u4e5f\u5927\u4e8e\u7b49\u4e8e\u96f6\u3002\u5982\u679cf\u52d2\u8d1d\u683c\u53ef\u79ef\u5e76\u4e14\u51e0\u4e4e\u603b\u662f\u5927\u4e8e\u7b49\u4e8e\u96f6\uff0c\u90a3\u4e48\u5b83\u7684\u52d2\u8d1d\u683c\u79ef\u5206\u4e5f\u5927\u4e8e\u7b49\u4e8e\u96f6\u3002
\u5e38\u89c1\u7684\u8d85\u8d8a\u79ef\u5206(\u4e0d\u53ef\u79ef\u79ef\u5206)
1\u3001\u222be^(ax^2)dx(a\u22600)
2\u3001\u222b(sinx)/xdx
3\u3001\u222b(cosx)/xdx
4\u3001\u222bsin(x^2)dx
5\u3001\u222bcos(x^2)dx
6\u3001\u222bx^n/lnxdx(n\u2260-1)
7\u3001\u222blnx/(x+a)dx(a\u22600)
8\u3001\u222b(sinx)^zdx(z\u4e0d\u662f\u6574\u6570)
\u6bd4\u5982\u222b[0,+\u221e)e^(-x^2)dx=\u221a\u03c0/2\uff0c\u6b64\u5904\u7684\u79ef\u5206\u503c\u5c31\u662f\u7528\u4e8c\u91cd\u79ef\u5206\u548c\u6781\u9650\u5939\u903c\u7684\u65b9\u6cd5\u5f97\u51fa\u7684\uff0c\u800c\u4e14\u53ea\u80fd\u7b97\u51fa(-\u221e,+\u221e)\u6216\u662f(0,+\u221e)\u4e0a\u7684\u503c\uff0c\u5176\u4ed6\u7684\u503c\u53ea\u80fd\u7528\u6570\u503c\u65b9\u6cd5\u7b97\u51fa\u8fd1\u4f3c\u503c\u3002
\u518d\u5982\u222b[0,+\u221e)(sinx)/xdx=\u03c0/2\uff0c\u6b64\u5904\u5c31\u662f\u7528\u7559\u6570\u7406\u8bba\u5f97\u51fa\u7684\u3002
\u4e00\u4e2a\u51fd\u6570\uff0c\u53ef\u4ee5\u5b58\u5728\u4e0d\u5b9a\u79ef\u5206\uff0c\u800c\u4e0d\u5b58\u5728\u5b9a\u79ef\u5206\uff0c\u4e5f\u53ef\u4ee5\u5b58\u5728\u5b9a\u79ef\u5206\uff0c\u800c\u6ca1\u6709\u4e0d\u5b9a\u79ef\u5206\u3002\u8fde\u7eed\u51fd\u6570\uff0c\u4e00\u5b9a\u5b58\u5728\u5b9a\u79ef\u5206\u548c\u4e0d\u5b9a\u79ef\u5206\u3002
\u82e5\u5728\u6709\u9650\u533a\u95f4[a,b]\u4e0a\u53ea\u6709\u6709\u9650\u4e2a\u95f4\u65ad\u70b9\u4e14\u51fd\u6570\u6709\u754c\uff0c\u5219\u5b9a\u79ef\u5206\u5b58\u5728\uff1b\u82e5\u6709\u8df3\u8dc3\u3001\u53ef\u53bb\u3001\u65e0\u7a77\u95f4\u65ad\u70b9\uff0c\u5219\u539f\u51fd\u6570\u4e00\u5b9a\u4e0d\u5b58\u5728\uff0c\u5373\u4e0d\u5b9a\u79ef\u5206\u4e00\u5b9a\u4e0d\u5b58\u5728\u3002
\u53c2\u8003\u8d44\u6599\u6765\u6e90\uff1a\u767e\u5ea6\u767e\u79d1\u2014\u2014\u4e0d\u5b9a\u79ef\u5206
sin x = x-x^3/3!+x^5/5!-...(-1)^(k-1)*x^(2k-1)/(2k-1)!+Rn(x)(-∞<x<∞)
所以：sinx/x=1-x^2/3!+x^4/5!-...(-1)^(k-1)*x^(2k-2)/(2k-1)!+Rn(x)(-∞<x<∞)
则∫sinx/xdx
=x-x^3/(3*3!)+x^5/(5*5!)-...(-1)^(k-1)*x^(2k-1)/(2k-1)*(2k-1)!)+Rn(x)(-∞<x<∞)
则其定积分为：
1-1/(3*3!)+1/(5*5!)-...(-1)^(k-1)/(2k-1)*(2k-1)!)+Rn(x)(-∞<x<∞)
可以求得一定的近似解
绛旓細x鈭圼2k蟺,(2k+1)蟺] k涓轰换鎰忔暣鏁帮紝鍘熷紡 = 鈭玸inx dx = - cosx + c x鈭(k蟺,(k+1)蟺) k涓轰换鎰忔暣鏁帮紝鍘熷紡 = -鈭玸inx dx = cosx + c 妤间笂鐨勨滃ぉ涔嬪敖_娴蜂箣婧愨 ,鐪嬫潵杩樺緱鍥炵倝锛岀湅闂澶偆娴呬簡锛屾妸棰樼洰鍋氭垚杩欐牱杩樼瑧鍒汉锛屼綘鑷繁鎶婄粨鏋滃啀寰垎涓娆＄湅鑳戒笉鑳藉緱鍒版ゼ涓荤粰鍑虹殑閭ｆ牱...
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绛旓細鈭(sinx)^4dx =鈭玔(1/2)(1-cos2x]^2dx =(1/4)鈭玔1-2cos2x+(cos2x)^2]dx =(1/4)鈭玔1-2cos2x+(1/2)(1+cos4x)]dx =(3/8)鈭玠x-(1/2)鈭玞os2xdx+(1/8)鈭玞os4xdx =(3/8)鈭玠x-(1/4)鈭玞os2xd2x+(1/32)鈭玞os4xd4x =(3/8)x-(1/4)sin2x+(1/32)sin4x+C...
绛旓細鈭玔0锛屜/2]sinxdx =-cosx[0锛屜/2]=1
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绛旓細鈭玿sinxdx =-鈭玿d(cosx)=-xcosx+鈭玞osxdx (搴旂敤鍒嗛儴绉垎娉)=-xcosx+sinx+C (C鏄Н鍒嗗父鏁)銆傚垎閮ㄧН鍒嗘硶鏄井绉垎瀛︿腑鐨勪竴绫婚噸瑕佺殑銆佸熀鏈殑璁＄畻绉垎鐨勬柟娉曘傚畠鏄敱寰垎鐨勪箻娉曟硶鍒欏拰寰Н鍒嗗熀鏈畾鐞嗘帹瀵艰屾潵鐨勩傚畠鐨勪富瑕佸師鐞嗘槸灏嗕笉鏄撶洿鎺ユ眰缁撴灉鐨勭Н鍒嗗舰寮忥紝杞寲涓虹瓑浠风殑鏄撴眰鍑虹粨鏋滅殑绉垎褰㈠紡鐨勩
绛旓細鍩烘湰鍏紡锛氣埆0dx=c锛涒埆x^udx=(x^u+1)/(u+1)+c锛涒埆1/xdx=ln|x|+c锛涒埆a^xdx=(a^x)/lna+c锛涒埆e^xdx=e^x+c锛鈭玸inxdx=-cosx+c锛涒埆cosxdx=sinx+c锛涒埆1/(cosx)^2dx=tanx+c锛涒埆1/(sinx)^2dx=-cotx+c 鎵╁睍鐭ヨ瘑锛氬湪鏁板涓紝骞虫柟鍙Н鍑芥暟鏄粷瀵瑰煎钩鏂圭殑绉垎涓烘湁闄愬肩殑瀹炲兼垨...
绛旓細鈭(sinx)^3 dx=(1/3)(cosx)^3 -cosx + C銆侰涓虹Н鍒嗗父鏁般傝В绛旇繃绋嬪涓嬶細鈭(sinx)^3 dx =-鈭(sinx)^2 dcosx =鈭玔(cosx)^2 -1 ] dcosx =(1/3)(cosx)^3 -cosx + C
绛旓細姹傝В杩囩▼濡備笅锛氳鈭玸inx/xdx=I锛屽垯锛欼=鈭埆{D}siny/ydxdy ,D鏄敱y=x锛寈=y^2鎵鍥存垚鐨勫钩闈㈠尯鍩熴傚埄鐢ㄥ垎閮ㄧН鍒嗘硶鏈夛細I=鈭珄0->1}siny/y (鈭珄y^2->y}dx)dy =鈭珄0->1}(siny/y) (y-y^2)dy =鈭珄0->1}(1-y)d[-cosy]=(1-1)[-cos1]-(1-0)d[-cos0]-鈭珄0->1}[-...
